Exercises

Find the derivative of $y=\cosh(x^2+1)$.

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$$y=\cosh(x^2+1)$$ $$y' = \sinh(x^2 + 1) \frac{d}{dx}(x^2 + 1)$$ $$y' = 2x\sinh(x^2 + 1)$$

Find the derivative of $y=\operatorname{csch}(e^x)$.

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$$y=\operatorname{csch}(e^x)$$ $$y' = -\operatorname{csch}x\coth x \frac{d}{dx}e^x$$ $$y' = -e^x \operatorname{csch}x\coth x$$

Find the derivative of $y=\operatorname{csch}(x+1) + \operatorname{sech}(x-1)$.

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$$y=\operatorname{csch}(x+1) + \operatorname{sech}(x-1)$$ $$y' = -\operatorname{csch}(x + 1) \coth (x + 1) \frac{d}{dx}(x-1) - \operatorname{sech}(x-1) \tanh (x-1) \frac{d}{dx}(x-1)$$ $$y' = -\operatorname{csch}(x + 1) \coth (x + 1) - \operatorname{sech}(x-1) \tanh (x-1)$$

Find the derivative of $y=\sqrt{2\tanh (x)}$.

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$$y=\sqrt{2\tanh (x)}$$ $$y=(2\tanh (x))^{1/2}$$ $$y' = \frac{1}{2} (2\tanh (x))^{-1/2} \frac{d}{dx}(2\tanh (x))$$ $$y' = \frac{1}{2(2\tanh (x))^{1/2}} (2\operatorname{sech}^2 x)$$ $$y' = \frac{\operatorname{sech}^2 x}{\sqrt{2\tanh (x)}}$$

Find the derivative of $y=(x+1)\coth x$.

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$$y=(x+1)\coth x$$ $$y'=(x+1)\frac{d}{dx}\coth x + \frac{d}{dx}(x+1) \cdot \coth x$$ $$y'=(x+1)(-\operatorname{csch}^2 x) + 1 \cdot \coth x$$ $$y'=-x\operatorname{csch}^2 x -\operatorname{csch}^2 x + \coth x$$

Find the derivative of $y=e^{\operatorname{sech} x}$.

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$$y=e^{\operatorname{sech} x}$$ $$y' = e^{\operatorname{sech}(x)} \frac{d}{dx} \operatorname{sech} x$$ $$y' = -e^{\operatorname{sech}(x)} \operatorname{sech} x \tanh x$$

Find $y'$ for the relation $\tanh(x+y)=x+y$.

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$$\tanh(x+y)=x+y$$ $$\operatorname{sech}(x + y)\frac{d}{dx}(x+y) = 1 + y'$$ $$\operatorname{sech}(x + y)(1+y') = 1 + y'$$

Resist the urge to divide both sides by $1 + y'$ here. It's generally a bad idea to divide both sides by something with the variable you're trying to solve for but, as you'll see in a minute, it's a particularly bad idea here.

$$\operatorname{sech}(x + y)+\operatorname{sech}(x + y)y' = 1 + y'$$ $$\operatorname{sech}(x + y)y'-y' = 1 - \operatorname{sech}(x + y)$$ $$y'(\operatorname{sech}(x + y)-1) = 1 - \operatorname{sech}(x + y)$$ $$y' = \frac{1 - \operatorname{sech}(x + y)}{\operatorname{sech}(x + y)-1}$$ $$y' = -1$$

This may seem like an odd result but, if you graph the original function, you'll see it's a diagonal line, the same as $y = -x$, so its slope is constant at -1.

This also shows why it would have been a very bad idea to divide both sides of the equation by $1 + y'$. Since $y' = -1$, that means that $1 + y' = 0$ which you can't divide an expression by.

Find $y'$ for the relation $x\operatorname{csch} y = x + 1$.

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$$x\operatorname{csch} y = x + 1$$ $$1 \cdot \operatorname{csch} y - x \operatorname{csch} y \coth y \cdot y' = 1$$ $$ - x \operatorname{csch} y \coth y \cdot y' = 1 - \operatorname{csch} y$$ $$ y' = \frac{-1 + \operatorname{csch} y}{x \operatorname{csch} y \coth y}$$

Find $y'$ for the relation $\cosh x + \sinh y = x + y$.

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$$\cosh x + \sinh y = x + y$$ $$\sinh x + \cosh y \cdot y' = 1 + y'$$ $$\cosh y \cdot y' - y'= 1 - \sinh x$$ $$y'(\cosh y - 1) = 1 - \sinh x$$ $$y' = \frac{1 - \sinh x}{\cosh y - 1}$$

Derive the formula for the derivative of $f(x) = \tanh x$. (Use the identity $\cosh^2 x - \sinh^2 x = 1$.)

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$$\begin{aligned} f(x) &= \tanh x \\ &= \frac{\sinh x}{\cosh x} \\ f'(x) &= \frac{\cosh x \frac{d}{dx} \sinh x - \sinh x \frac{d}{dx} \cosh x}{\cosh^2 x} \\ &= \frac{\cosh x \cosh x - \sinh x \sinh x}{\cosh^2 x} \\ &= \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x} \\ &= \frac{1}{\cosh^2 x} \\ &= \operatorname{sech}^2 x \end{aligned}$$

Derive the formula for the derivative of $f(x) = \operatorname{sech} x$.

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Rewrite the function as $f(x) = (\cosh x)^{-1}$ then apply the Chain Rule.