Question

How do you find the derivative of ${{\sinh }^{4}}(x)$?

Answer 1

We are given an expression which we have to differentiate. We will be using the chain rule for the same. Firstly, we will differentiate the power in the expression, so we will get $4{{\sinh }^{3}}(x)$. Then, we will differentiate the base trigonometric function which is $\sinh (x)$. Hence, we will have derivatives of the given expression.

Complete step by step answer:
According to the given question, we are given an expression whose derivative we have to find. And for that, we will be using the chain rule.
But first, we will first know what is $\sinh (x)$. The $\sinh (x)$ is the hyperbolic sine function. It can be said as the analogue of the sine function $\sin (x)$ that we usually use in the trigonometry. But both of them are different.
Sine function, $\sin (x)$is used in a right angled triangle in which $x$ is an angle. And we can write that,
$\sin (x)=\dfrac{perpendicular}{hypotenuse}$
whereas $\sinh (x)$ is the hyperbolic sine of $x$ and we can write that,
$\sinh (x)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$
There is another one which is, $\arcsin$ and this refers to the inverse sine function.

The given expression we have is,
${{\sinh }^{4}}(x)$----(1)
Differentiating the equation (1), we get,
$\dfrac{d}{dx}({{\sinh }^{4}}(x))$
First, the powers on the base function will get differentiated and we have,
$\Rightarrow 4{{\sinh }^{3}}(x).\dfrac{d}{dx}(\sinh (x))$
Now, we will differentiate the base function involved, and we get,
$\Rightarrow 4{{\sinh }^{3}}(x).\cosh (x)$

Therefore, the derivative of ${{\sinh }^{4}}(x)$ is $4{{\sinh }^{3}}(x).\cosh (x)$.

Note: The differentiation of the above question proceeded step wise and using chain rule. $\sin (x)$ and $\sinh (x)$ are two different trigonometric functions and should not be mistaken as misspelt sine functions.