Product rule for Differentiation

Covid-19 has led the world to go through a phenomenal transition .

E-learning is the future today.

Stay Home , Stay Safe and keep learning!!!

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.

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

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

$\frac{\text{d}}{\text{d}x}[u.v] = v.\frac{\text{d}u}{\text{d}x} + u.\frac{\text{d}v}{\text{d}x}$

$\frac{\text{d}}{\text{d}x}[u.v] = v.u ' + u.v '$

Note : Order of the function does not matter as addition is a commutative property.

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

Home

Covid-19 has affected physical interactions between people.

Don't let it affect your learning.