Question

What is the derivative of $\sinh \left( x \right)$?

Answer 1

Hint: We must first express the hyperbolic function $\sinh \left( x \right)$ in exponential terms. We know that $\sinh \left( x \right)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}$, and then we can differentiate this function in exponential terms to get the required answer using the definition of hyperbolic cosine, $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.

Complete step by step solution:
We know that as the trigonometric functions relate to a circle, the hyperbolic functions relate to a hyperbola. $\sinh \left( x \right)$ and $\cosh \left( x \right)$ are two such hyperbolic functions.
We know that $\sinh \left( x \right)$ is called the hyperbolic sine function. It is defined as
$\sinh \left( x \right)=\dfrac{{{e}^{x}}-{{e}^{-x}}}{2}...\left( i \right)$
Similarly, $\cosh \left( x \right)$ is called the hyperbolic cosine function and we can define it as
$\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}...\left( ii \right)$
In this question, we need to find the derivative of $\sinh \left( x \right)$, that is, $\dfrac{d}{dx}\sinh \left( x \right)$.
Using the equation (i), we can write
$\dfrac{d}{dx}\sinh \left( x \right)=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}-{{e}^{-x}}}{2} \right)$
We can also write this as
$\dfrac{d}{dx}\sinh \left( x \right)=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{2}-\dfrac{{{e}^{-x}}}{2} \right)$
Now, we know the property that $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$.
Hence, we can write
$\dfrac{d}{dx}\sinh \left( x \right)=\dfrac{d}{dx}\left( \dfrac{{{e}^{x}}}{2} \right)-\dfrac{d}{dx}\left( \dfrac{{{e}^{-x}}}{2} \right)...\left( iii \right)$
We know that the differentiation of ${{e}^{x}}$, with respect to x is ${{e}^{x}}$, that is, $\dfrac{d}{dx}\left( {{e}^{x}} \right)={{e}^{x}}...\left( iv \right)$
And by using the chain rule of differentiation, we can write
$\dfrac{d}{dx}\left( {{e}^{-x}} \right)={{e}^{-x}}\times \left( -1 \right)$, that is, $\dfrac{d}{dx}\left( {{e}^{-x}} \right)=-{{e}^{-x}}...\left( v \right)$
Now, by using the equations (iv) and (v) and the property of multiplication by a constant in differentiation, we can write equation (iii) as
$\dfrac{d}{dx}\sinh \left( x \right)=\left( \dfrac{{{e}^{x}}}{2} \right)-\left( -\dfrac{{{e}^{-x}}}{2} \right)$
Thus, we can also write the above equation as
$\dfrac{d}{dx}\sinh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$
From equation (ii), we know that $\cosh \left( x \right)=\dfrac{{{e}^{x}}+{{e}^{-x}}}{2}$.
Thus, we can write,
$\dfrac{d}{dx}\sinh \left( x \right)=\cosh \left( x \right)$.
Hence, we can say that the derivative of $\sinh \left( x \right)$ is $\cosh \left( x \right)$.

Note: We must understand that the sine hyperbolic function is a different function than the sine function. Also, we must see the question carefully and not consider $\sinh \left( x \right)$ as $\sin \left( hx \right)$. And since $\sinh \left( x \right)$ is a different function, its graph will also be different from that of $\sin \left( x \right)$.