Math Doubts

Proof of differentiation of Hyperbolic Sine function

$x$ is a variable and the hyperbolic sine is written as $\sinh{x}$ in mathematical form. The differentiation of $\sinh{x}$ with respect to $x$ is written as $\dfrac{d}{dx}{\, \sinh{x}}$ in mathematical form. The derivative of hyperbolic sine function with respect to $x$ can be expressed in limit form by the definition of a derivative.

$\dfrac{d}{dx}{\, f(x)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{ \normalsize \dfrac{f(x+h)-f(x)}{h}}$

Take $f(x) \,=\, \sinh{x}$, then $f{(x+h)} \,=\, \sinh{(x+h)}$. Substitute them to start deriving the derivative of the hyperbolic sine function with respect to $x$.

$\implies \dfrac{d}{dx}{\, \sinh{x}}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\sinh{(x+h)}-\sinh{x}}{h}}$

Express Hyperbolic sine in exponential form

The hyperbolic sine functions $\sinh{x}$ and $\sinh{(x+h)}$ can be written in terms of exponential terms as per the fundamental definition of hyperbolic sine.

$\sinh{x} \,=\, \dfrac{e^x-e^{-x}}{2}$

$\sinh{(x+h)} \,=\, \dfrac{e^{(x+h)}-e^{-(x+h)}}{2}$

It is time to substitute expansions of $\sinh{x}$ and $\sinh{(x+h)}$ functions in the definition of derivative of sin function.

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}-e^{-(x+h)}}{2}-\dfrac{e^x-e^{-x}}{2}}{h}}$

Simplify the exponential form expression

The functions in the numerator is in complex form. So, try to simplify it mathematically. It allows us to find the derivative of hyperbolic sine function in next few steps.

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}-e^{-(x+h)}-e^x+e^{-x}}{2}}{h}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{\dfrac{e^{(x+h)}-e^{-(x+h)}-e^x+e^{-x}}{2}}{\dfrac{h}{1}}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{(x+h)}-e^{-(x+h)}-e^x+e^{-x}}{2} \times \dfrac{1}{h}}$

Now, multiply the fractions to simplify it further.

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{[e^{(x+h)}-e^{-(x+h)}-e^x+e^{-x}] \times 1}{2 \times h}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{(x+h)}-e^{-(x+h)}-e^x+e^{-x}}{2h}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x} \times e^{h}-e^{-x-h}-e^x+e^{-x}}{2h}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}e^{h}-e^{-x}e^{-h}-e^{x}+e^{-x}}{2h}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}e^{h}-e^{x}-e^{-x}e^{-h}+e^{-x}}{2h}}$

$e^x$ is a common factor in the first two terms of the numerator and $e^{-x}$ is another common factor in the third and fourth terms of the numerator.

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(-e^{-h}+1)}}{2h}}$

$= \,\,\,$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{1}{2} \times \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(-e^{-h}+1)}}{h}}$

$= \,\,\,$ $\dfrac{1}{2} \times \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(-e^{-h}+1)}}{h}}$

$= \,\,\,$ $\dfrac{1}{2} \large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}+e^{-x}{(-e^{-h}+1)}}{h}}$

$= \,\,\,$ $\dfrac{1}{2} \large \displaystyle \lim_{h \to 0}{\normalsize \Bigg[\dfrac{e^{x}{(e^{h}-1)}}{h}+\dfrac{e^{-x}{(-e^{-h}+1)}}{h}}\Bigg]$

According to sum rule of limits, the limit of sum of functions can be written as sum of their limits.

$= \,\,\,$ $\dfrac{1}{2} \Bigg[\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}}{h}}$ $+$ $\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-x}{(-e^{-h}+1)}}{h}}\Bigg]$

$= \,\,\,$ $\dfrac{1}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{x}{(e^{h}-1)}}{h}}$ $+$ $\dfrac{1}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-x}{(-e^{-h}+1)}}{h}}$

$e^x$ is a common function in the first term and $e^{-x}$ is also a common function in the second term of the expression because the limit of each function has to be evaluated as $h$ approaches $0$.

$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{-e^{-h}+1}{h}}$

Find the limit of each function

The mathematical expression is successfully simplified and it is time to find the limit of each function.

$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{-(e^{-h}-1)}{h}}$

$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{-h}-1}{-h}}$

$= \,\,\,$ $\dfrac{e^{x}}{2}\large \displaystyle \lim_{h \to 0}{\normalsize \dfrac{e^{h}-1}{h}}$ $+$ $\dfrac{e^{-x}}{2}\large \displaystyle \lim_{-h \to 0}{\normalsize \dfrac{e^{-h}-1}{-h}}$

As per exponential limit rule, the limit of each function is equal to one.

$= \,\,\,$ $\dfrac{e^{x}}{2} \times 1$ $+$ $\dfrac{e^{-x}}{2} \times 1$

$= \,\,\,$ $\dfrac{e^{x}}{2} + \dfrac{e^{-x}}{2}$

$= \,\,\,$ $\dfrac{e^{x}+e^{-x}}{2}$

$= \,\,\,$ $\cosh{x}$

$\therefore \,\,\, \dfrac{d}{dx}{\, \sinh{x}} \,=\, \cosh{x}$

Therefore, it is proved that the derivative of hyperbolic sine function with respective to $x$ is equal to hyperbolic cosine.



Follow us
Email subscription
Math Doubts
Math Doubts is a best place to learn mathematics and from basics to advanced scientific level for students, teachers and researchers. Know more
Follow us on Social Media
Mobile App for Android users Math Doubts Android App
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more