Differentiating inverse trigonometric and hyperbolic functionsEdexcel International A Level Further Maths: Revision notes

Three results are in the formulae booklet, and you should know them: $\frac{d}{dx}\arcsin x=\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}\arccos x=-\frac{1}{\sqrt{1-x^2}},\quad\frac{d}{dx}\arctan x=\frac{1}{1+x^2}.$ To prove one, let $y=\arcsin x$, so $x=\sin y$. Then $\frac{dx}{dy}=\cos y=\sqrt{1-\sin^2y}=\sqrt{1-x^2}$ and $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}$. The positive root is taken because $y$ lies in $\left[-\frac\pi2,\frac\pi2\right]$, where $\cos y\geq0$. The derivative of $\arcsin x$ exists only for $|x|<1$, where the gradient of the curve is finite.

$\frac{d}{dx}\operatorname{arsinh}x=\frac{1}{\sqrt{x^2+1}},\quad\frac{d}{dx}\operatorname{arcosh}x=\frac{1}{\sqrt{x^2-1}}\ (x>1),\quad\frac{d}{dx}\operatorname{artanh}x=\frac{1}{1-x^2}\ (|x|<1).$ Proof for $\operatorname{arsinh}$: $x=\sinh y$ gives $\frac{dx}{dy}=\cosh y=\sqrt{1+\sinh^2y}=\sqrt{1+x^2}$, so $\frac{dy}{dx}=\frac{1}{\sqrt{1+x^2}}$. Compare the sign patterns with the trigonometric results. The $\arcsin$ result has $1-x^2$ under the root, $\operatorname{arsinh}$ has $x^2+1$ and $\operatorname{arcosh}$ has $x^2-1$. The gradient of $\operatorname{arcosh}x$ is undefined at $x=1$, where the curve is vertical.

When the argument is a function $g(x)$, multiply by $g'(x)$: $\frac{d}{dx}\arcsin(kx)=\frac{k}{\sqrt{1-k^2x^2}},\qquad\frac{d}{dx}\arctan\left(x^2\right)=\frac{2x}{1+x^4}.$ Example: $\frac{d}{dx}\left(\frac12\operatorname{artanh}x^2\right)=\frac12\cdot\frac{2x}{1-x^4}=\frac{x}{1-x^4}$. Example: $\frac{d}{dx}\operatorname{arcosh}(2x)=\frac{2}{\sqrt{4x^2-1}}$. Write the substitution out first: replace $x$ in the standard result by the whole argument, then multiply by its derivative.

Many questions combine inverse functions with products and roots, and the answer often simplifies. $\frac{d}{dx}\left(\arcsin x+x\sqrt{1-x^2}\right)=\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}-\frac{x^2}{\sqrt{1-x^2}}=\frac{2(1-x^2)}{\sqrt{1-x^2}}=2\sqrt{1-x^2}.$ $\frac{d}{dx}\left(x\operatorname{arsinh}x-\sqrt{1+x^2}\right)=\operatorname{arsinh}x+\frac{x}{\sqrt{1+x^2}}-\frac{x}{\sqrt{1+x^2}}=\operatorname{arsinh}x.$ Put terms over a common denominator, and look for cancellation. The same results are used later in reverse to integrate inverse functions. For tangents and normals, evaluate the function and its derivative at the given $x$. Exact values: $\arcsin\frac12=\frac\pi6$, $\arctan1=\frac\pi4$, and $\operatorname{arsinh}x=\ln\left(x+\sqrt{x^2+1}\right)$.