# How to find the derivative of a function of a function. Please give an example.

### 2 Answers | Add Yours

To find the derivative of the function of a function.

Solution:

To find the derivative y = (x^2 )^3. Or y = f (g(x) ). Where g(x) = x^2 and f(g(x)) = (x^2)^3

We know that d/dx f((g(x)) = (df/dg)(dg/dx).

df/dg = 3(g(x)}^2-1 = 3(x^2)^3-1 = 3(x^2 )^2 = 3x^4.

dg/dx = d/dx(x^2) = 2x.

df(g(x))/ dx = 3(x^4 )(2x) = 6x^5....(1)

Verification:

Actually d/dx (x^2)^3 = d/dx { x^6} , (x^2)^3 = x^6.

d/dx (x^6 ) = 6 * x^(6-1) = 6x^5.....(2).

Since (1) and (2) are same , we differentiated the function (x^2)^3 which is in the form f(g(x)) and verified also.

Examle 2:

y = sin( ax+b)).

This in the form y = f(g(x). Where g(x) ax^n+b and f(g(x) = sin (ax^n+b).

we diferentiate the function sin (ax+b) with respect to ax+b and then mutiply by (ax+b)'.

y = sin(ax+b)

dy/dx = y' = {sin(ax+b)}'

y' = {cos(ax+b)}*(ax+b)'

y' = {cos(ax+b)}(a)

y' = a cos (ax+b).

Now we

To determine the derivative of a function of a function, we'll have to apply the chain rule.

If u is a function of v and v is a function of x, then we can say that u is the function of the function v.

We'll write in this manner:

du/dx = (du/dv)*(dv/dx)

The first step is to find the function v (usually is the function inside the brackets or the argument of the trigonometric functions, or the argument of the logarithmic functions, or the expression under the square root, etc.)

Then, we'll re-write the function u in terms of v and we'll differentiate u with respect to v.

We'll re-write the results with respect to x.

Example:

u = (x^2 + 5x)^3

We'll have to differentiate u with respect to x:

du/dx = (d/dx)(x^2 + 5x)^3

We'll substitute the expression inside the brackets by v.

v = x^2 + 5x

u = v^3

To apply the chain rule, we'll have to differentiate u with respect to v:

du/dv = (v^3)'

du/dv = 3v^2

Now, we'll differentiate v with respect to x:

dv/dx = (x^2 + 5x)'

dv/dx = 2x + 5

du/dx = (du/dv)(dv/dx)

**du/dx = 3[(x^2 + 5x)^2]*(2x + 5)**