FP3.3 - Differentiation

Hyperbolic derivatives follow from the exponential definitions and then obey the usual chain, product, quotient and reciprocal rules. The key pair mirrors differentiation of $e^x$:$$\frac{d}{dx}\sinh x=\cosh x,\qquad \frac{d}{dx}\cosh x=\sinh x.$$Unlike trigonometric differentiation, differentiating $\cosh x$ does not introduce a minus sign.

For example, the quotient rule and $\cosh^2u-\sinh^2u=1$ give$$\frac{d}{du}\tanh u=\frac{\cosh^2u-\sinh^2u}{\cosh^2u}=\operatorname{sech}^2u.$$The reciprocal rule then gives the negative sign in $d(\operatorname{sech}u)/du=-\operatorname{sech}u\tanh u$. These derivations are safer than importing trigonometric signs by memory.

Differentiate the outside function, keep its argument, then multiply by the derivative of that argument. Thus$$\frac{d}{dx}\tanh(5x-1)=5\operatorname{sech}^2(5x-1),$$and$$\frac{d}{dx}\operatorname{sech}(x^2)=-2x\operatorname{sech}(x^2)\tanh(x^2).$$Every nested layer contributes a factor.

For a product such as $y=x\sinh^2x$, keep both product-rule terms and apply a second chain rule to the square:$$\frac{dy}{dx}=\sinh^2x+2x\sinh x\cosh x=\sinh^2x+x\sinh2x.$$The identity $2\sinh x\cosh x=\sinh2x$ is an optional final simplification, not a replacement for the differentiation steps.

Write powers explicitly before differentiating a quotient. If$$y=\frac{\cosh2x}{\sqrt{x+1}}=\cosh(2x)(x+1)^{-1/2},\qquad x>-1,$$then$$\frac{dy}{dx}=2\sinh(2x)(x+1)^{-1/2}-\frac12\cosh(2x)(x+1)^{-3/2}.$$A common denominator may be formed only after both terms are correct.

If $y=f^{-1}(x)$ on a one-to-one branch of $f$, then $x=f(y)$ and implicit differentiation gives$$\frac{dy}{dx}=\frac{1}{f'(y)}.$$Rewrite the result in terms of $x$ using the defining identity and the principal range. The branch and domain are part of the derivative formula, especially when a square root appears.

To derive the inverse-sinh result, let $y=\operatorname{arsinh}x$, so $x=\sinh y$. Then$$1=\cosh y\frac{dy}{dx}\quad\Longrightarrow\quad\frac{dy}{dx}=\frac1{\cosh y}.$$Because $\cosh y>0$ and $\cosh^2y=1+\sinh^2y$, this becomes$$\frac{d}{dx}\operatorname{arsinh}x=\frac1{\sqrt{1+x^2}}.$$The positive square root is justified by the principal range, not chosen arbitrarily.

For a composite inverse function, differentiate the complete inner expression. For example, if $y=\operatorname{arcosh}(2x+3)$ and $x>-1$, then$$\frac{dy}{dx}=\frac{2}{\sqrt{(2x+3)^2-1}}.$$The stricter condition $2x+3>1$ ensures the derivative is finite; the function itself is defined at the endpoint where $2x+3=1$, but its derivative formula is singular there.

Useful simplification often appears only after applying the rule. For$$y=\arcsin x+x\sqrt{1-x^2},\qquad |x|<1,$$the product and chain rules give$$\frac{dy}{dx}=\frac1{\sqrt{1-x^2}}+\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=2\sqrt{1-x^2}.$$Keeping a common radical visible makes the cancellation reliable.

When simplifying radicals, use $\sqrt{x^2}=|x|$, not automatically $x$. For example,$$\frac{d}{dx}\operatorname{arsinh}\!\left(\sqrt{4x^2-1}\right)=\frac{2x}{|x|\sqrt{4x^2-1}},\qquad |x|>\frac12.$$This equals $2/\sqrt{4x^2-1}$ for $x>1/2$ but $-2/\sqrt{4x^2-1}$ for $x<-1/2$. A stated domain decides the sign.

After differentiating a combination, solve $f'(x)=0$ within the original domain. If clearing radicals requires squaring, the resulting equation is only a candidate equation: substitute each candidate back into the unsquared derivative equation and reject values outside the inverse-function or radical domain.