Derivative of sinh x with respect to x

Formula

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

Proof

$x$ is a literal number and represents a real number. The hyperbolic function is written as $\sinh{x}$ mathematically and differentiation of the hyperbolic sine function with respect to $x$ is expressed in mathematical form as follows.

$\dfrac{d}{dx} \, \sinh{x}$

01

Differentiation of a function in Limit form

According to the differentiation of a function in limit form, the differentiation of a function $f(x)$ can be written as follows.

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

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

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

02

Transformation of Hyperbolic sine function in exponential form

Each hyperbolic sine function can be expressed in terms of exponential functions. So, write both hyperbolic sine functions in exponential form.

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

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

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

03

Simplifying the exponential expression

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

An expression is formed in the numerator and it is in terms of exponential functions. The first two exponential functions can be written in product form of two exponential functions as per product rule of exponents.

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

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

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

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

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

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

04

Simplifying the function by the Limit Rules

The limit value belongs to both functions. Hence, it can be applied to both functions by the Limit rule of addition of functions.

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

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

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

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

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

Limit of the each function is in the form of the limit of (e^x-1)/x as x approaches 0 formula. As per this rule, the value of limit of each function is one.

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

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

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

The ratio of sum of the $e$ raised to the power of $x$ and $e$ raised to the power of $–x$ to the number $2$ is hyperbolic cosine function and it is written as $\cosh{x}$ in mathematical form.

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

Therefore, it is proved that the derivative of hyperbolic sine function $\sinh{x}$ with respect to $x$ is hyperbolic cosine function $\cosh{x}$.


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