Question

If $f(x)=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{6}}}{6!}+........\infty$, then show that $f'(x)=\sinh x$.

Answer 1

Hint:We will first differentiate the given expression $f(x)=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{6}}}{6!}+........\infty$ then expand the factorial using the formula $n!=n(n-1)!$ and transform the given expression into simplest form by cancelling similar terms and also we should know about expansion formula of $\,\sinh x$.

Complete step-by-step answer:
Before proceeding with the question, we should understand the concept behind differentiation and also about hyperbolic sine function. Differentiation allows us to find rates of change. For example, it allows us to find the rate of change of velocity with respect to time (which is acceleration). It also allows us to find the rate of change of x with respect to y, which on a graph of y against x is the gradient of the curve. There are a number of simple rules which can be used to allow us to differentiate many functions easily.
Sine hyperbolic function can be defined as simple rational functions of the exponential function of z: $\sinh z=\dfrac{{{e}^{z}}-{{e}^{-z}}}{2}$.
If y = some function of x (in other words if y is equal to an expression containing numbers and x's), then the derivative of y (with respect to x) is written $\dfrac{dy}{dx}$.
$\,f(x)=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{6}}}{6!}+........\infty \,.......(1)$
Differentiation of x to the power something: If $y={{x}^{n}}$ then $\dfrac{dy}{dx}=n\cdot {{x}^{n-1}}.........(2)$
Differentiating equation (1) by using information from equation (2) we get,
$f'(x)=0+\dfrac{2x}{2!}+\dfrac{4{{x}^{3}}}{4!}+\dfrac{6{{x}^{5}}}{6!}+........\infty \,...\,...(3)$
Simplifying equation (3) and using $n!=1\times 2\times 3.......\times n$ we get,
$f'(x)=0+x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+........\infty \,\,...........(4)$
Now equation (4) is expansion of $\sinh x$ i.e $\sinh x=x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+........\infty$.
So,$f'(x)=\sinh x$ Hence proved.

Note: Students should know the differentiation formulas and factorial(n!) formula, this will take less time in solving the question.They need to be careful while cancelling the terms from numerator and denominator by expanding the factorial.Also students should remember expansion of sine hyperbolic function $\sinh x=x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+........\infty$ and cos hyperbolic function $\cosh x=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+........\infty$ for solving these types of questions.From above question we can say that derivative of hyperbolic function coshx is sinhx as expansion of coshx is given above.