Product rule for Differentiation

The product rule for differentiation of two functions 'f' and 'g' in which both the functions should be differentiable. Moreover, the derivative of is the first function times the derivative of the second, plus the second function times the derivative of the first.

Prove that : $\frac{\text{d}}{\text{d}x}[f(x)g(x)] =g(x) \frac{\text{d}}{\text{d}x}f(x) + f(x)\frac{\text{d}}{\text{d}x}g(x)$

Proof : $\frac{\text{d}}{\text{d}x}[f(x)g(x)] =\lim_{\triangle x \rightarrow 0} \frac{f(x + \triangle x)g(x +\triangle x)- f(x)g(x)}{\triangle x}$

Add and subtract $f(x +\triangle x)g(x)$

$ { =\lim_{\triangle x \rightarrow 0} \frac{f(x + \triangle x)g(x +\triangle x)\color{red}{-f(x +\triangle x)g(x) +f(x +\triangle x)g(x)}- f(x)g(x)}{\triangle x}}$

=$\lim_{\triangle x \rightarrow 0}[f(x +\triangle x)\frac{g(x + \triangle x) -g(x)}{\triangle x}+ g(x)\frac{f(x + \triangle x) - f(x)}{\triangle x}]$

=$\lim_{\triangle x \rightarrow 0}[f(x +\triangle x)\frac{g(x + \triangle x) -g(x)}{\triangle x}]+ \lim_{\triangle x \rightarrow 0}[ g(x)\frac{f(x + \triangle x) - f(x)}{\triangle x}]$

=$\lim_{\triangle x \rightarrow 0}[f(x +\triangle x)\lim_{\triangle x \rightarrow 0}\frac{g(x + \triangle x) -g(x)}{\triangle x}]+ \lim_{\triangle x \rightarrow 0}g(x)\lim_{\triangle x \rightarrow 0}[\frac{f(x + \triangle x) - f(x)}{\triangle x}]$

By the definition of derivative and $\lim_{\triangle x \rightarrow 0}[f(x +\triangle x)$ = f(x)
$\frac{\text{d}}{\text{d}x}[f(x)g(x)] = f(x)g'(x) + g(x)f'(x)$

If there are are more than two functions then product rule can be extended.

$\frac{\text{d}}{\text{d}x}[f(x)g(x)h(x)] = f(x).h(x)g'(x) + g(x)h(x)f'(x) + f(x)g(x)h'(x)$

Examples on product rule for differentiation

1) Find the derivative of f(x)= $x^{3}$cos(x)
Solution : f(x)= $x^{3}$cos(x)
Using the product rule,

$\frac{\text{d}}{\text{d}x}[f(x)g(x)] =g(x) \frac{\text{d}}{\text{d}x}f(x) + f(x)\frac{\text{d}}{\text{d}x}g(x)$

f(x) = $x^{3}$ and g(x) = cos(x)

$\frac{\text{d}}{\text{d}x}[x^{3} cos(x)] =cos(x) \frac{\text{d}}{\text{d}x}x^{3} + x^{3}\frac{\text{d}}{\text{d}x}cos(x)$

= $cos(x).3x^{2} + x^{3}(-sin(x))$ [ using the power rule and derivative of trigonometric function]

$\frac{\text{d}}{\text{d}x}[x^{3} cos(x)] = 3cos(x).x^{2} - x^{3}sin(x)$

2) Find the derivative of f(x)= $\sqrt{x}tan(x)$
Solution : f(x)= $\sqrt{x}tan(x)$
Using the product rule,

$\frac{\text{d}}{\text{d}x}[f(x)g(x)] =g(x) \frac{\text{d}}{\text{d}x}f(x) + f(x)\frac{\text{d}}{\text{d}x}g(x)$

f(x) = $\sqrt{x}$ and g(x) = tan(x)

$\frac{\text{d}}{\text{d}x}[\sqrt{x}tan(x)] =tan(x) \frac{\text{d}}{\text{d}x}\sqrt{x} + \sqrt{x}\frac{\text{d}}{\text{d}x}tan(x)$

= $tan(x).\frac{1}{2}(x)^{\frac{-1}{2}} + \sqrt{x}sec{2}x$ [ using the power rule and derivative of trigonometric function]

=$tan(x).\frac{1}{2\sqrt{x}}+\sqrt{x}sec{2}x$

$\frac{\text{d}}{\text{d}x}[\sqrt{x}.tan(x)] = tan(x).\frac{1}{2\sqrt{x}}+\sqrt{x}.sec^{2}x$

3) Find the derivative of f(x) = x.cos(x) and find f'(c) at c= $\frac{\pi}{4}$

Solution : f(x) = x.cos(x)
Using the product rule,

$\frac{\text{d}}{\text{d}x}[f(x)g(x)] =g(x) \frac{\text{d}}{\text{d}x}f(x) + f(x)\frac{\text{d}}{\text{d}x}g(x)$

f(x) = x and g(x) = cos(x)

$\frac{\text{d}}{\text{d}x}[x. cos(x)] =cos(x). \frac{\text{d}}{\text{d}x}x + x.\frac{\text{d}}{\text{d}x}cos(x)$

= cos(x).(1) + x.(-sin(x)) [ using the power rule and derivative of trigonometric function]

f '(x) = cos(x) - x.sin(x)

f '(c) = cos(c) - c.sin(c)

When c = $\frac{\pi}{4}$

$f'(\frac{\pi}{4}) = cos(\frac{\pi}{4}) -\frac{\pi}{4}.sin(\frac{\pi}{4})$

$f'(\frac{\pi}{4}) = cos(\frac{\sqrt{2}}{2}) -\frac{\pi}{4}.sin(\frac{\sqrt{2}}{2})$

12th grade math

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