Question

How do you find the derivative of $y = {e^{\cosh \left( {2x} \right)}}$?

Answer 1

Hint: First find the differentiation of $2x$ with respect to $x$. Then, find the differentiation of $\cosh \left( {2x} \right)$ with respect to $2x$. Then, find the differentiation of ${e^{\cosh \left( {2x} \right)}}$ with respect to $\cosh \left( {2x} \right)$. Multiply these and use chain rule to get the required derivative.

Complete step by step solution:
We have to find the derivative of $y = {e^{\cosh \left( {2x} \right)}}$.
Here, $f\left( x \right) = {e^{g\left( x \right)}}$, where $g\left( x \right) = \cosh \left( {h\left( x \right)} \right)$ and $h\left( x \right) = 2x$.
We have to find the differentiation of $f$ with respect to $x$.
It can be done using Chain Rule.
$\dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dh}} \times \dfrac{{dh}}{{dx}}$……(1)
i.e., Differentiation of $f$ with respect to $x$ is equal to product of differentiation of $f$ with respect to $g$, and differentiation of $g$ with respect to $h$, and differentiation of $h$ with respect to $x$.
We will first find the differentiation of $h$ with respect to $x$.
Here, $h\left( x \right) = 2x$
Differentiating $h$ with respect to $x$.
$\dfrac{{dh}}{{dx}} = \dfrac{d}{{dx}}\left( {2x} \right)$
Now, using the property that the differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
So, in above differentiation, constant $2$ can be taken outside the differentiation.
$ \Rightarrow \dfrac{{dh}}{{dx}} = 2\dfrac{d}{{dx}}\left( x \right)$
Now, using the differentiation formula $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$ in above differentiation, we get
$ \Rightarrow \dfrac{{dh}}{{dx}} = 2$……(2)
Now, we will find the differentiation of$g$ with respect to $h$.
Here, $g\left( x \right) = \cosh \left( {h\left( x \right)} \right)$
Differentiating$g$ with respect to $h$.
$\dfrac{{dg}}{{dh}} = \dfrac{d}{{dh}}\left( {\cosh \left( {h\left( x \right)} \right)} \right)$
The derivative of the cosh function is $\dfrac{d}{{dx}}\left( {\cosh x} \right) = \sinh x$.
$ \Rightarrow \dfrac{{dg}}{{dh}} = \sinh \left( {h\left( x \right)} \right)$
Put the value of $h\left( x \right)$ in the above equation.
Since, $h\left( x \right) = 2x$
So, $\dfrac{{dg}}{{dh}} = \sinh \left( {2x} \right)$……(3)
Now, we will find the differentiation of$f$ with respect to $g$.
Here, $f\left( x \right) = {e^{g\left( x \right)}}$
Differentiating$f$ with respect to $g$.
$\dfrac{{df}}{{dg}} = \dfrac{d}{{dg}}\left( {{e^{g\left( x \right)}}} \right)$
The derivative of exponential function is $\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}$.
$ \Rightarrow \dfrac{{df}}{{dg}} = {e^{g\left( x \right)}}$
Put the value of $g\left( x \right)$ in the above equation.
Since, $g\left( x \right) = \cosh \left( {h\left( x \right)} \right)$ and $h\left( x \right) = 2x$
So, $\dfrac{{df}}{{dg}} = {e^{\cosh \left( {2x} \right)}}$…….(4)
Put the value of $\dfrac{{df}}{{dg}},\dfrac{{dg}}{{dh}},\dfrac{{dh}}{{dx}}$ from Equation (2), (3) and (4) in Equation (1).
$\dfrac{{df}}{{dx}} = {e^{\cosh \left( {2x} \right)}} \times \sinh \left( {2x} \right) \times 2$
Multiplying the terms, we get
$ \Rightarrow \dfrac{{df}}{{dx}} = 2\sinh \left( {2x} \right){e^{\cosh \left( {2x} \right)}}$

Therefore, the derivative of $y = {e^{\cosh \left( {2x} \right)}}$ is $y' = 2\sinh \left( {2x} \right){e^{\cosh \left( {2x} \right)}}$.

Note: Chain rule, in calculus, basic method for differentiating a composite function. If $f\left( x \right)$ and $g\left( x \right)$ are two functions, the function $f\left( {g\left( x \right)} \right)$ is calculated for a value of $x$ by first evaluating $g\left( x \right)$ and then evaluating the function $f$ at this value of $g\left( x \right)$, thus “chaining” the results together.