Question

Find the derivative of $y = {\sinh ^{ - 1}}(2x)$ ?

Answer 1

Calculate the derivative of the function $y = {\sinh ^{ - 1}}(2x)$.
Apply the chain rule to solve this question,
Use chain rule : Consider $y = f(u)$ and $u = g(x)$ are differentiable functions. Then is derivative is given by, $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Take $y = {\sinh ^{ - 1}}u$ and $u = 2x$ , then find derivatives $\dfrac{{dy}}{{du}}$ and $\dfrac{{du}}{{dx}}$.
Use the derivative of the inverse hyperbolic formulas ; $\dfrac{d}{{dx}}{\sinh ^{ - 1}}x = \dfrac{1}{{\sqrt {1 + {x^2}} }}$ .

Complete step by step answer:
Consider the function $y = {\sinh ^{ - 1}}(2x)$.
Apply the Chain rule: Consider $y = f(u)$ and $u = g(x)$ are differentiable functions. Then is derivative is given by,
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$
Here, $y = {\sinh ^{ - 1}}u$ and $u = 2x$.
Find derivative of $y = {\sinh ^{ - 1}}u$,
$\dfrac{{d\left( {{{\sinh }^{ - 1}}u} \right)}}{{du}} = \dfrac{1}{{\sqrt {1 + {u^2}} }} \ldots (1)$
Find derivative of $u = 2x$,
$\dfrac{{d\left( {2x} \right)}}{{dx}} = 2 \ldots (2)$
The derivative of $x$ is $1$ .
Multiply equation $(1)$ and equation $(2)$.
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 + {u^2}} }} \times 2$
$\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 + {u^2}} }}$
Substitute $u = 2x$ into the derivative$\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 + {u^2}} }}$.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 + {{(2x)}^2}} }}$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 + 4{x^2}} }}$
Final Answer: The derivative of $y = {\sinh ^{ - 1}}(2x)$ is $\dfrac{{dy}}{{dx}} = \dfrac{2}{{\sqrt {1 + 4{x^2}} }}$.

Note: Find the list of formulas of derivative of he hyperbolic function,
$\dfrac{d}{{dx}}(\sinh x) = \cosh x$
$\dfrac{d}{{dx}}(\cosh x) = \sinh x$
$\dfrac{d}{{dx}}(\tanh x) = {\operatorname{sech} ^2}x$
$\dfrac{d}{{dx}}{\sinh ^{ - 1}}x = \dfrac{1}{{\sqrt {1 + {x^2}} }}$
$\dfrac{d}{{dx}}{\cosh ^{ - 1}}x = \dfrac{1}{{\sqrt {{x^2} - 1} }}$
$\dfrac{d}{{dx}}{\tanh ^{ - 1}}x = \dfrac{1}{{1 - {x^2}}}$