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$$\sinh(x) = \frac{1}{2(e^x - e^{-x})}$$ $$\cosh(x) = \frac{1}{2(e^x + e^{-x}}$$ $$\tanh(x) = \frac{\sinh (x)}{\cosh (x)}$$

Prove: $$\frac{d(\tanh(x))}{dx} = \frac{1}{(\cosh x)^2}$$

I got the derivative for $\tanh(x)$ as: $$\left[ \frac{1}{2(e^x + e^{-x})}\right]^2 - \frac{{[ \frac{1}{2(e^x + e^{-x})}]^2}}{[ \frac{1}{2(e^x + e^{-x})}]^2}$$

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    $\begingroup$ Welcome to math.SE. Please read about proper typesetting in LaTeX and make sure your questions meet the general standards of the site - you will get used to the norm by reading a few of the questions. $\endgroup$ – Andrew Thompson Apr 5 '14 at 17:03
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$$\dfrac{d(f/g)}{dx}=\dfrac{gf^\prime-fg^\prime}{g^2}$$ Set $f=\sinh,g=\cosh$ to get $$\dfrac{d\tanh}{dx}=\dfrac{\cosh\cdot\sinh^\prime-\sinh\cdot\cosh^\prime}{\cosh^2}$$ Now, $$\sinh^\prime=\dfrac{1}{2}(e^x+e^{-x})=\cosh\\ \cosh^\prime=\dfrac{1}{2}(e^x-e^{-x})=\sinh$$ Thus, $$\dfrac{d\tanh}{dx}=\dfrac{\cosh^2-\sinh^2}{\cosh^2}=1-\left(\dfrac{\sinh}{\cosh}\right)^2=1-\tanh^2$$ Now, $$\dfrac{1}{\cosh^2}=1-\tanh^2$$ (Proof: $$1-\dfrac{\sinh^2 x}{\cosh^2 x}= \dfrac{\cosh^2 x-\sinh^2 x}{\cosh^2 x}$$ Since $(\cosh^2 x) - (\sinh^2 x) = 1$, $$\dfrac{1}{\cosh^2 x} = {\operatorname{sech}^2 x} $$) Thus, $$\boxed{\dfrac{d\tanh}{dx}=1-\tanh^2=\dfrac{1}{\cosh^2 x}}$$

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Observe that $$(a+b)^2-(a-b)^2=4ab$$

Set $a=e^x,b=e^{-x}$ and $\displaystyle \operatorname{sech}x=\frac1{\cosh x}=\frac2{e^x+e^{-x}}$

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