hcju/proofwikiqa huggingface dataset:
{{begin-eqn}} {{eqn | l = \laptrans {e^{i a t} }       | r = \frac 1 {s - i a}       | c = Laplace Transform of Exponential }} {{eqn | r = \frac {s + i a} {s^2 + a^2}       | c = multiply top and bottom by $s + i a$ }}  {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \laptrans {e^{i a t} }       | r = \laptrans {\cos a t + i \sin a t}       | c = Euler's Formula }} {{eqn | r = \laptrans {\cos a t} + i \laptrans {\sin a t}       | c = Linear Combination of Laplace Transforms }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \laptrans {\cos a t}       | r = \map \Re {\laptrans {e^{i a t} } } }} {{eqn | r = \map \Re {\frac {s + i a} {s^2 + a^2} } }} {{eqn | r = \frac s {s^2 + a^2} }} {{end-eqn}} {{qed}}
By definition of the Laplace Transform: :$\displaystyle \laptrans {\cos at} = \int_0^{\to +\infty} e^{-s t} \cos at \rd t$ From Integration by Parts: :$\displaystyle \int f g' \rd t = f g - \int f'g \rd t$ Here: {{begin-eqn}} {{eqn | l = f       | r = \cos at }} {{eqn | ll= \leadsto       | l = f'       | r = -a \sin a t       | c = Derivative of Cosine Function }} {{eqn | l = g'       | r = e^{-s t} }} {{eqn | ll= \leadsto       | l = g       | r = -\frac 1 s e^{-s t}       | c = Primitive of Exponential Function }} {{end-eqn}} So: {{begin-eqn}} {{eqn | n = 1       | l = \int e^{-s t} \cos a t \rd t       | r = -\frac 1 s e^{-s t} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t }} {{end-eqn}} Consider: :$\displaystyle \int e^{-s t} \sin a t \rd t$ Again, using Integration by Parts: :$\displaystyle \int h j \,' \rd t = h j - \int h'j \rd t$ Here: {{begin-eqn}} {{eqn | l = h       | r = \sin at }} {{eqn | ll= \leadsto       | l = h'       | r = a \cos at       | c = Derivative of Sine Function }} {{eqn | l = j\,'       | r = e^{-s t} }} {{eqn | ll= \leadsto       | l = j       | r = -\frac 1 s e^{-s t}       | c = Primitive of Exponential Function }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \int e^{-s t} \sin a t \rd t       | r = -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t }} {{end-eqn}} Substituting this into $(1)$: {{begin-eqn}} {{eqn | l = \int e^{-s t} \cos a t \rd t       | r = -\frac 1 s e^{-s t} \cos a t - \frac a s \paren {-\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t} }} {{eqn | r = -\frac 1 s e^{-s t} \cos a t + \frac a {s^2} e^{-s t} \sin a - \frac {a^2} {s^2} \int e^{-s t} \cos a t \rd t }} {{eqn | ll= \leadsto       | l = \paren {1 + \frac {a^2} {s^2} } \int e^{-s t} \cos a t \rd t       | r = -\frac 1 s e^{-s t} \cos a t + \frac a {s^2} e^{-s t} \sin a t }} {{end-eqn}} Evaluating at $t = 0$ and $t \to +\infty$: {{begin-eqn}} {{eqn | l = \paren {1 + \frac {a^2} {s^2} } \laptrans {\cos at}       | r = \intlimits {-e^{-s t} \paren {\frac 1 s \cos a t - \frac a {s^2} \sin a t} } {t \mathop = 0} {t \mathop \to +\infty} }} {{eqn | r = 0 - \paren {-1 \paren {\frac 1 s \times 1 + \frac a {s^2} \times 0} }       | c = Boundedness of Real Sine and Cosine, Complex Exponential Tends to Zero }} {{eqn | r = \frac 1 s }} {{eqn | ll= \leadsto       | l = \laptrans {\cos at}       | r = \frac 1 s \paren {1 + \frac {a^2} {s^2} }^{-1} }} {{eqn | r = \frac 1 s \paren {\frac {s^2} {a^2 + s^2} } }} {{eqn | r = \frac s {s^2 + a^2} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\cos a t}       | r = \laptrans {\frac {e^{i a t} + e^{-i a t} } 2}       | c = Cosine Exponential Formulation }} {{eqn | r = \frac 1 2 \paren {\laptrans {e^{i a t} } + \laptrans {e^{-i a t} } }       | c = Linear Combination of Laplace Transforms }} {{eqn | r = \frac 1 2 \paren {\frac 1 {s - i a} + \frac 1 {s + i a} }       | c = Laplace Transform of Exponential }} {{eqn | r = \frac 1 2 \paren {\frac {s + i a + s - i a} {s^2 + a^2} }       | c = simplifying }} {{eqn | r = \frac s {s^2 + a^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\laptrans {\cos {a t} } } s       | r = \int_0^{\to +\infty} e^{-s t} \cos {a t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \cos {a t} \rd t       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{L \mathop \to \infty} \intlimits {\frac {e^{-s t} \paren {-s \cos a t + a \sin a t} } {\paren {-s}^2 + a^2} } 0 L       | c = Primitive of $e^{a x} \cos b x$ }} {{eqn | r = \lim_{L \mathop \to \infty} \paren {\frac {e^{-s L} \paren {-s \cos a L + a \sin a L} } {s^2 + a^2} - \frac {e^{-s \times 0} \paren {-s \, \map \cos {0 \times a} + a \, \map \sin {0 \times a} } } {s^2 + a^2} }       | c =  }} {{eqn | r = \lim_{L \mathop \to \infty} \paren {\frac {s \, \map \cos {0 \times a} - a \, \map \sin {0 \times a} } {s^2 + a^2} - \frac {e^{-s L} \paren {-s \cos a L + a \sin a L} } {s^2 + a^2} }       | c =  }} {{eqn | r = \frac {s \, \map \cos {0 \times a} - a \, \map \sin {0 \times a} } {s^2 + a^2} - 0       | c = Exponential Tends to Zero }} {{eqn | r = \frac {s \cos 0 - a \sin 0} {s^2 + a^2}       | c = simplifying }} {{eqn | r = \frac s {s^2 + a^2}       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{end-eqn}} {{qed}}
Let $\map f t := \map \Ci t = \displaystyle \int_t^\infty \dfrac {\cos u} u \rd u$. Then: {{begin-eqn}} {{eqn | l = \map {f'} t       | r = -\dfrac {\cos t} t       | c =  }} {{eqn | ll= \leadsto       | l = t \map {f'} t       | r = -\cos t       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {t \map {f'} t}       | r = -\laptrans {\cos t}       | c =  }} {{eqn | r = -\dfrac s {s^2 + 1}       | c = Laplace Transform of Cosine }} {{eqn | ll= \leadsto       | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}       | r = -\dfrac s {s^2 + 1}       | c = Derivative of Laplace Transform }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}       | r = \dfrac s {s^2 + 1}       | c = Laplace Transform of Derivative }} {{eqn | ll= \leadsto       | l = s \laptrans {\map f t}       | r = \int \dfrac s {s^2 + 1} \rd s       | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$ }} {{eqn | ll= \leadsto       | l = s \laptrans {\map f t}       | r = \dfrac 1 2 \map \ln {s^2 + 1} + C       | c = Primitive of $\dfrac x {x^2 + a^2}$ }} {{end-eqn}} By the Initial Value Theorem of Laplace Transform: :$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$ which leads to: :$c = 0$ Thus: {{begin-eqn}} {{eqn | l = s \laptrans {\map f t}       | r = \dfrac 1 2 \map \ln {s^2 + 1}       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\map f t}       | r = \dfrac {\map \ln {s^2 + 1} } {2 s}       | c =  }} {{end-eqn}} {{qed}}
Let $\map f t = \sin \sqrt t$. Then: {{begin-eqn}} {{eqn | l = \map {f'} t       | r = \dfrac {\cos \sqrt t} {2 \sqrt t}       | c =  }} {{eqn | l = \map f 0       | r = 0       | c =  }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \laptrans {\map {f'} t}       | r = \dfrac 1 2 \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }       | c =  }} {{eqn | r = s \, \map F s - \map f 0       | c = Laplace Transform of Derivative }} {{eqn | r = \dfrac {\sqrt \pi} {2 s^{1/2} } \map \exp {-\dfrac 1 {4 s} }       | c = Laplace Transform of Sine of Root }} {{eqn | ll= \leadsto       | l = \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }       | r = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {e^{i a t} }       | r = \frac 1 {s - i a}       | c = Laplace Transform of Exponential }} {{eqn | r = \frac {s + i a} {s^2 + a^2}       | c = multiplying top and bottom by $s + i a$ }}  {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \laptrans {e^{i a t} }       | r = \laptrans {\cos a t + i \sin a t}       | c = Euler's Formula }} {{eqn | r = \laptrans {\cos a t} + i \laptrans {\sin a t}       | c = Linear Combination of Laplace Transforms }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \laptrans {\sin a t}       | r = \map \Im {\laptrans {e^{i a t} } } }} {{eqn | r = \map \Im {\frac {s + i a} {s^2 + a^2} } }} {{eqn | r = \frac a {s^2 + a^2} }} {{end-eqn}} {{qed}}
By definition of the Laplace transform: :$\displaystyle \laptrans {\sin a t} = \int_0^{\to +\infty} e^{-s t} \sin a t \rd t$ From Integration by Parts: :$\displaystyle \int f g' \rd t = f g - \int f' g \rd t$ Here: {{begin-eqn}} {{eqn | l = f       | r = \sin a t }} {{eqn | ll= \leadsto       | l = f'       | r = a \cos a t       | c = Derivative of $\sin a x$ }} {{eqn | l = g'       | r = e^{-s t} }} {{eqn | ll= \leadsto       | l = g       | r = -\frac 1 s e^{-s t}       | c = Primitive of $e^{a x}$ }} {{end-eqn}} So: {{begin-eqn}} {{eqn | n = 1       | l = \int e^{-s t} \sin a t \rd t       | r = -\frac 1 s e^{-s t} \sin a t + \frac a s \int e^{-s t} \cos a t \rd t }} {{end-eqn}} Consider: :$\displaystyle \int e^{-s t} \cos a t \rd t$ Again, using Integration by Parts: :$\displaystyle \int h j\,' \rd t = h j - \int h' j \rd t$ Here: {{begin-eqn}} {{eqn | l = h       | r = \cos a t }} {{eqn | ll= \leadsto       | l = h'       | r = -a \sin a t       | c = Derivative of Cosine Function }} {{eqn | l = j\,'       | r = e^{-s t} }} {{eqn | ll= \leadsto       | l = j       | r = -\frac 1 s e^{-s t}       | c = Primitive of Exponential Function }} {{end-eqn}} So: {{begin-eqn}} {{eqn | l = \int e^{-s t} \cos a t \rd t       | r = -\frac 1 s e^{-st} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t }} {{end-eqn}} Substituting this into $(1)$: {{begin-eqn}} {{eqn | l = \int e^{-s t} \sin a t \rd t       | r = -\frac 1 s e^{-s t} \sin a t + \frac a s \paren {-\frac 1 s e^{-s t} \cos a t - \frac a s \int e^{-s t} \sin a t \rd t} }} {{eqn | r = -\frac 1 s e^{-s t} \sin a t - \frac a {s^2} e^{-s t} \cos a t - \frac {a^2} {s^2} \int e^{-s t} \sin a t \rd t }} {{eqn | ll= \leadsto       | l = \paren {1 + \frac {a^2} {s^2} } \int e^{-s t} \sin a t \rd t       | r = -e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t} }} {{end-eqn}} Evaluating at $t = 0$ and $t \to +\infty$: {{begin-eqn}} {{eqn | l = \paren {1 + \frac {a^2} {s^2} } \laptrans {\sin a t}        | r = \intlimits {-e^{-s t} \paren {\frac 1 s \sin a t + \frac a {s^2} \cos a t} } {t \mathop = 0} {t \mathop \to +\infty} }} {{eqn | r = 0 - \paren {-1 \paren {\frac 1 s \times 0 + \frac a {s^2} \times 1} }       | c = Boundedness of Real Sine and Cosine, Complex Exponential Tends to Zero }} {{eqn | r = \frac a {s^2} }} {{eqn | ll= \leadsto       | l = \laptrans {\sin a t}       | r = \frac a {s^2} \paren {1 + \frac {a^2} {s^2} }^{-1} }} {{eqn | r = \frac a {s^2} \paren {\frac {s^2} {a^2 + s^2} } }} {{eqn | r = \frac a {s^2 + a^2} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\sin a t}       | r = \laptrans {\frac {e^{i a t} - e^{-i a t} } {2 i} }       | c = Sine Exponential Formulation }} {{eqn | r = \frac 1 {2 i} \paren {\laptrans {e^{i a t} } - \laptrans {e^{-i a t} } }       | c = Linear Combination of Laplace Transforms }} {{eqn | r = \frac 1 {2 i} \paren {\frac 1 {s - i a} - \frac 1 {s + i a} }       | c = Laplace Transform of Exponential }} {{eqn | r = \frac 1 {2 i} \paren {\frac {s + i a - s + i a} {s^2 + a^2} }       | c = simplifying }} {{eqn | r = \frac 1 {2 i} \paren {\frac {2 i a} {s^2 + a^2} }       | c = simplifying }} {{eqn | r = \frac a {s^2 + a^2}       | c = simplifying }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map {\laptrans {\sin {a t} } } s       | r = \int_0^{\to +\infty} e^{-s t} \sin {a t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \lim_{L \mathop \to \infty} \int_0^L e^{-s t} \sin {a t} \rd t       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{L \mathop \to \infty} \intlimits {\frac {e^{-s t} \paren {-s \sin a t - a \cos a t} } {\paren {-s}^2 + a^2} } 0 L       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | r = \lim_{L \mathop \to \infty} \paren {\dfrac {e^{-s L} \paren {-s \sin a L - a \cos a L} } {s^2 + a^2} - \dfrac {e^{-s \times 0} \paren {-s \, \map \sin {0 \times a} - a \, \map \cos {0 \times a} } } {s^2 + a^2} }       | c =  }} {{eqn | r = \lim_{L \mathop \to \infty} \paren {\dfrac {s \sin 0 + a \cos 0} {s^2 + a^2} - \dfrac {e^{-s L} \paren {s \sin a L + a \cos a L} } {s^2 + a^2} }       | c = Exponential of Zero and rearranging }} {{eqn | r = \dfrac {s \sin 0 + a \cos 0} {s^2 + a^2} - 0       | c = Exponential Tends to Zero }} {{eqn | r = \frac a {s^2 + a^2}       | c = Sine of Zero is Zero, Cosine of Zero is One }} {{end-eqn}} {{qed}}
Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$. Then: :$\map f 0 = 0$ and: {{begin-eqn}} {{eqn | l = \map {f'} t       | r = \dfrac {\sin t} t       | c =  }} {{eqn | ll= \leadsto       | l = t \map {f'} t       | r = \sin t       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {t \map {f'} t}       | r = \laptrans {\sin t}       | c =  }} {{eqn | r = \dfrac 1 {s^2 + 1}       | c = Laplace Transform of Sine }} {{eqn | ll= \leadsto       | l = -\dfrac \d {\d s} \laptrans {\map {f'} t}       | r = \dfrac 1 {s^2 + 1}       | c = Derivative of Laplace Transform }} {{eqn | ll= \leadsto       | l = \map {\dfrac \d {\d s} } {s \laptrans {\map f t} - \map f 0}       | r = -\dfrac 1 {s^2 + 1}       | c = Laplace Transform of Derivative }} {{eqn | ll= \leadsto       | l = s \laptrans {\map f t}       | r = -\int \dfrac 1 {s^2 + 1} \rd s       | c = $\map f 0 = 0$, and integrating both sides {{WRT|Integration}} $s$ }} {{eqn | ll= \leadsto       | l = s \laptrans {\map f t}       | r = -\arctan s + C       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{end-eqn}} By the Initial Value Theorem of Laplace Transform: :$\displaystyle \lim_{s \mathop \to \infty} s \laptrans {\map f t} = \lim_{t \mathop \to 0} \map f t = \map f 0 = 0$ which leads to: :$c = \dfrac \pi 2$ Thus: {{begin-eqn}} {{eqn | l = s \laptrans {\map f t}       | r = \dfrac \pi 2 - \arctan s       | c =  }} {{eqn | r = \arccot s       | c = Sum of Arctangent and Arccotangent }} {{eqn | r = \arctan \dfrac 1 s       | c = Arctangent of Reciprocal equals Arccotangent }} {{eqn | ll= \leadsto       | l = \laptrans {\map f t}       | r = \dfrac 1 s \arctan \dfrac 1 s       | c =  }} {{end-eqn}} {{qed}}
Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$. Then: :$\map f 0 = 0$ and: {{begin-eqn}} {{eqn | l = \map \Si t       | r = \int_0^t \dfrac {\sin u} u \rd u       | c = {{Defof|Sine Integral Function}} }} {{eqn | r = \int_0^t \dfrac 1 u \paren {u - \dfrac {u^3} {3!} + \dfrac {u^5} {5!} - \dfrac {u^7} {7!} + \dotsb} \rd u       | c = {{Defof|Real Sine Function}} }} {{eqn | r = t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb       | c = Primitive of Power }} {{eqn | ll= \leadsto       | l = \laptrans {\map \Si t}       | r = \laptrans {t - \dfrac {t^3} {3 \times 3!} + \dfrac {t^5} {5 \times 5!} - \dfrac {t^7} {7 \times 7!} + \dotsb}       | c =  }} {{eqn | r = \dfrac 1 {s^2} - \dfrac 1 {3 \times 3!} \dfrac {3!} {s^4} + \dfrac 1 {5 \times 5!} \dfrac {5!} {s^6} - \dfrac 1 {7 \times 7!} \dfrac {7!} {s^8} + \dotsb       | c = Laplace Transform of Positive Integer Power }} {{eqn | r = \dfrac 1 {s^2} - \dfrac 1 {3 s^4} + \dfrac 1 {5 s^6} - \dfrac 1 {7 s^8} + \dotsb       | c = simplifying }} {{eqn | r = \dfrac 1 s \paren {\dfrac {\paren {1 / s} } 1 - \dfrac {\paren {1 / s}^3} 3 + \dfrac {\paren {1 / s}^5} 5 - \dfrac {\paren {1 / s}^7} 7 + \dotsb}       | c = rearranging }} {{eqn | r = \dfrac 1 s \arctan \dfrac 1 s       | c = Power Series Expansion for Real Arctangent Function }} {{end-eqn}} {{qed}}
Let $\map f t := \map \Si t = \displaystyle \int_0^t \dfrac {\sin u} u \rd u$. Then: :$\map f 0 = 0$ and: {{begin-eqn}} {{eqn | l = \map \Si t       | r = \int_0^t \dfrac {\sin u} u \rd u       | c = {{Defof|Sine Integral Function}} }} {{eqn | r = \int_0^1 \dfrac {\sin t v} v \rd v       | c = Integration by Substitution $u = t v$ }} {{eqn | ll= \leadsto       | l = \laptrans {\map \Si t}       | r = \laptrans {\int_0^1 \dfrac {\sin t v} v \rd v}       | c =  }} {{eqn | r = \int_0^\infty e^{-s t} \paren {\int_0^1 \dfrac {\sin t v} v \rd v} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^1 \dfrac 1 v \paren {\int_0^\infty e^{-s t} \sin t v \rd t} \rd v       | c = exchanging order of integration }} {{eqn | r = \int_0^1 \dfrac {\laptrans {\sin t v} } v \rd v       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^1 \dfrac {\d v} {s^2 + v^2}       | c = Laplace Transform of Sine }} {{eqn | r = \intlimits {\dfrac 1 s \arctan \dfrac v s} 0 1       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \dfrac 1 s \arctan \dfrac 1 s       | c =  }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \sin \sqrt t       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1}!}       | c = {{Defof|Real Sine Function}} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}!} t^{n + \frac 1 2}       | c =  }} {{eqn | ll= \leadsto       | l = \laptrans {\sin \sqrt t}       | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1}! s^{n + \frac 3 2} }       | c = Laplace Transform of Power, Linear Combination of Laplace Transforms }} {{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1}! s^n}       | c = Gamma Difference Equation }} {{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {2 n + 1} \paren {2 n}!} {2^{2 n} n! \paren {2 n + 1}! s^n}       | c = Gamma Function of Positive Half-Integer }} {{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {2^{2 n} n! s^n}       | c =  }} {{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {-\frac 1 {2^2 s} }^n       | c =  }} {{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } e^{-1/\paren {2^2 s} }       | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}} }} {{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }       | c = simplifying }} {{end-eqn}} {{qed}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition: :$\displaystyle D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$ $P \left({1}\right)$ is true, as this just says: :$D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$ which is trivially true. === Basis for the Induction === $P \left({2}\right)$ is the case: :$\displaystyle D_x \left({f_1 \left({x}\right) + f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) + D_x  \left({f_2 \left({x}\right)}\right)$ which has been proved in Sum Rule for Derivatives. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true. So this is the induction hypothesis: :$\displaystyle D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^k D_x \left({f_i \left({x}\right)}\right)$ from which it is to be shown that: :$\displaystyle D_x \left({\sum_{i \mathop = 1}^{k + 1} f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^{k + 1} D_x \left({f_i \left({x}\right)}\right)$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = D_x \left({\sum_{i \mathop = 1}^{k + 1} f_i \left({x}\right)}\right)       | r = D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right) + f_{k+1} \left({x}\right)}\right)       | c = Definition of Summation }} {{eqn | r = D_x \left({\sum_{i \mathop = 1}^k f_i \left({x}\right)}\right) + D_x \left({f_{k+1} \left({x}\right)}\right)       | c = Basis of the Induction }} {{eqn | r = \sum_{i \mathop = 1}^k D_x \left({f_i \left({x}\right)}\right) + D_x \left({f_{k+1} \left({x}\right)}\right)       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k + 1} D_x \left({f_i \left({x}\right)}\right)       | c = Definition of Summation }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \N_{>0}: D_x \left({\sum_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n D_x \left({f_i \left({x}\right)}\right)$ {{qed}}
Proof by induction: For all $n \in \N_{\ge 1}$, let $P \left({n}\right)$ be the proposition: : $\displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$ $P(1)$ is true, as this just says: : $D_x \left({f_1 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right)$ === Basis for the Induction === $P(2)$ is the case: : $D_x \left({f_1 \left({x}\right) f_2 \left({x}\right)}\right) = D_x \left({f_1 \left({x}\right)}\right) f_2 \left({x}\right) + f_1 \left({x}\right) D_x \left({f_2 \left({x}\right)}\right)$ which has been proved in Product Rule for Derivatives. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true. So this is our induction hypothesis: :$\displaystyle D_x \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$ Then we need to show: :$\displaystyle D_x \left({\prod_{i \mathop = 1}^{k+1} f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = D_x \left({\prod_{i \mathop = 1}^{k+1} f_i \left({x}\right)}\right)       | r = D_x \left({\left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)}\right)       | c =  }} {{eqn | r = D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) + D_x \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) f_{k+1} \left({x}\right)       | c = Basis for the Induction }} {{eqn | r = D_x \left({f_{k+1} \left({x}\right)}\right) \left({\prod_{i \mathop = 1}^k f_i \left({x}\right)}\right) + \left({\sum_{i \mathop = 1}^k \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)}\right) f_{k+1} \left({x}\right)       | c = Induction Hypothesis }} {{eqn | r = \sum_{i \mathop = 1}^{k+1} \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle D_x \left({\prod_{i \mathop = 1}^n f_i \left({x}\right)}\right) = \sum_{i \mathop = 1}^n \left({D_x \left({f_i \left({x}\right)}\right) \prod_{j \mathop \ne i} f_j \left({x}\right)}\right)$ for all $n \in \N$ {{qed}}
{{begin-eqn}} {{eqn | l = y       | r = \arccot x       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \cot y       | c = {{Defof|Real Arccotangent}} }} {{eqn | ll= \leadsto       | l = \frac {\d x} {\d y}       | r = -\csc^2 y       | c = Derivative of Cotangent Function }} {{eqn | r = -\paren {1 + \cot^2 y}       | c = Difference of Squares of Cosecant and Cotangent }} {{eqn | r = -\paren {1 + x^2}       | c = Definition of $x$ }} {{eqn | ll= \leadsto       | l = \frac {\d y} {\d x}       | r = \frac {-1} {1 + x^2}       | c = Derivative of Inverse Function }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \frac {\map \d {\arccot x} } {\d x}       | r = \map {\frac \d {\d x} } {\frac \pi 2 - \arctan x}       | c = Tangent of Complement equals Cotangent }} {{eqn | r = -\frac 1 {1 + x^2}       | c = Derivative of Arctangent Function }} {{end-eqn}} {{qed}}