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. **

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

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)= $\sqrt{x}tan(x)$

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

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

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

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

Home

Russia-Ukraine crisis update - 3rd Mar 2022

The UN General assembly voted at an emergency session to demand an immediate halt to Moscow's attack on Ukraine and withdrawal of Russian troops.

GMAT

GRE

1st Grade

2nd Grade

3rd Grade

4th Grade

5th Grade

6th Grade

7th grade math

8th grade math

9th grade math

10th grade math

11th grade math

12th grade math

Precalculus

Worksheets

Chapter wise Test

MCQ's

Math Dictionary

Graph Dictionary

Multiplicative tables

Math Teasers

NTSE

Chinese Numbers

CBSE Sample Papers