Calculus of hyperbolic functionsEdexcel A-Level Further Maths: Revision notes

Differentiate the definitions term by term, using $\frac{d}{dx}e^{-x}=-e^{-x}$: $\frac{d}{dx}\sinh x=\cosh x,\qquad\frac{d}{dx}\cosh x=\sinh x,\qquad\frac{d}{dx}\tanh x=\frac{1}{\cosh^2x}=\operatorname{sech}^2x.$ Unlike $\cos x$, differentiating $\cosh x$ gives $+\sinh x$, with no minus sign. The result for $\tanh x$ follows from the quotient rule with $\cosh^2x-\sinh^2x=1$.

The usual rules apply with these derivatives.

• Chain rule: $\frac{d}{dx}\tanh3x=\frac{3}{\cosh^23x}=3\operatorname{sech}^23x$, and $\frac{d}{dx}\cosh3x=3\sinh3x$.
• Product rule: $\frac{d}{dx}\left(x\sinh^2x\right)=\sinh^2x+2x\sinh x\cosh x$. Exact values at $x=\ln a$ use $e^{\ln a}=a$: $\sinh(\ln3)=\frac43$, $\cosh(\ln3)=\frac53$. So the gradient of $y=x\sinh^2x$ at $x=\ln3$ is $\frac{16}{9}+\frac{40}{9}\ln3$.

$\int\sinh x\,dx=\cosh x+c,\quad\int\cosh x\,dx=\sinh x+c,\quad\int\frac{1}{\cosh^2x}\,dx=\tanh x+c.$ For a linear argument, divide by the coefficient: $\int\cosh kx\,dx=\frac1k\sinh kx+c$ and $\int\sinh kx\,dx=\frac1k\cosh kx+c$. Remember the constant $c$ for an indefinite integral. For a definite integral, give the exact value using $e$ or $\ln$ where possible.

Look for an integrand of the form $\frac{f'(x)}{\sqrt{f(x)}}$, or a function times its derivative, and recognise it as the result of the chain rule. Example: $\frac{d}{dx}\sqrt{1+\sinh2x}=\frac{2\cosh2x}{2\sqrt{1+\sinh2x}}=\frac{\cosh2x}{\sqrt{1+\sinh2x}}$, so $\int\frac{\cosh2x}{\sqrt{1+\sinh2x}}\,dx=\sqrt{1+\sinh2x}+c.$ Always differentiate your answer to check it. Similarly $\int\sinh x\cosh x\,dx=\frac12\sinh^2x+c$.

• Stationary points: solve $\frac{dy}{dx}=0$. If this gives $\tanh x=k$ with $|k|\ge1$, there are no stationary points.
• Tangent at a point: find the gradient from the derivative, then $y-y_1=m(x-x_1)$.
• Areas: integrate between limits, checking the curve is on one side of the axis. Find where a curve meets the $x$-axis by writing $\sinh$ and $\cosh$ as exponentials. Example: $y=2\cosh x-3\sinh x$ has $\frac{dy}{dx}=2\sinh x-3\cosh x$, which is never $0$ because $\tanh x=\frac32$ is impossible; the area from $0$ to $\ln2$ is $\frac34$.