# Product rule for Differentiation

We at ask-math believe that educational material should be free for everyone. Please use the content of this website for in-depth understanding of the concepts. Additionally, we have created and posted videos on our youtube.

We also offer One to One / Group Tutoring sessions / Homework help for Mathematics from Grade 4th to 12th for algebra, geometry, trigonometry, pre-calculus, and calculus for US, UK, Europe, South east Asia and UAE students.

Affiliations with Schools & Educational institutions are also welcome.

Please reach out to us on [email protected] / Whatsapp +919998367796 / Skype id: anitagovilkar.abhijit

We will be happy to post videos as per your requirements also. Do write to us.

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

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

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