Derivatives of Composite Functions (original) (raw)

Last Updated : 22 Nov, 2025

Derivative of Composite Functions is the derivative of the function that is composite. A composite function is a function in which the variable is itself a function of the variable. The derivative of these functions is easily found using the derivative of the composite function formula that is explained below in the article.

Derivatives of Composite Functions Formula

Suppose we have a composite function h(x) that is represented as, h(x) = f{g(x)}, then first we find the derivative of f(x) and multiply the same with the derivative of g(x) to get the required derivative of the composite function.

d/dx.f{g(x)} = f'(g(x)).g'(x)

Composite Functions and Chain Rule

Let's say we have a function f(x) = (x + 1)2, for which we want to calculate the derivative. These kinds of functions are called composite functions, which means they are made up of more than one function. Usually, they are of the form g(x) = h(f(x)) or it can also be written as g = hof(x). In our case, the given function f(x) = (x + 1)2 is composed of two functions,

f(x) = g(h(x))

where,

**For example,

f(x) = (x + 1)2

f(x) = x2 + 1 + 2x

Differentiating the function with respect to x,

f'(x) = 2x + 1

**Chain Rule

Let f be a real-valued function which is a composite of two functions, “u” and “v”, that is f = v o u. Let's say t = u(x) and if both dv/dtanddt/dxexist for both of the functions "u" and "v".

dv/dx = (dv/dt).(dt/dx)

The chain rule can be extended to any number of composite functions. For example,

f = (w o u) o v

If t = v(x) and s = u(t)

Then,

\frac{df}{dx} = \frac{d[w (u(v))]}{dt}\cdot \frac{dt}{dx} = \frac{dw}{ds}\cdot\frac{ds}{dt}\cdot\frac{dt}{dx}

Let's say we have a function f(x) = sin(x2)

This function is a composite function made up of two functions. If t = u(x) = x2 and v(t) = sin(t), then

f(x) = (v o u)(x)

= v(u(x))

= v(x2)

= sin x2

Putting t = u(t) = x2

\frac{dv}{dt} = cos(t) and \frac{dt}{dx} = 2x

Hence, by chain rule,

\frac{df}{dx} = \frac{dv}{dt}\cdot\frac{dt}{dx} = cos(t).2x

\frac{df}{dx} = cos(x^2).2x

**Alternative Method to Chain Rule

The chain rule can also be applied with a shortcut method. This is explained with an example, let's say we have a function f(x) = (sin(x)) 2

In general, we don't really use the composition of the functions approach to differentiate the functions. We identify the “inside function” and the “outside function”. Then, differentiate the outside function leaving the inside function alone, and keep going in this manner.

**For example, Differentiate sin 2 x

df/dx = d/dx{sin2x}

df/dx = 2sin x{d/dx(sin x)}

df/dx = 2.(sin x).(cos x)

Derivatives of Composite Functions In One Variable

Using the chain rule the derivative of composite function in one variable is easily find. This is explained as by the example, find the derivative of (x2 + 3)3

= d/dx(x2 + 3)3

Using Chain Rule

= 3(x2 + 3)2.d/dx(x2 + 3)

= 3(x2 + 3)2.(2x)

= 6x(x2 + 3)2

Derivatives of Composite Functions Examples

**Examples 1: Find the derivative for the function f(x) = (x + 1)2.

**Solution:

df/dx = d/dx(x + 1)2

df/dx = 2(x + 1)d/dx(x + 1)

df/dx = 2(x + 1).1

df/dx = 2x + 2

**Examples 3: Find the derivative for the function f(x) = (x2 + 1)5

**Solution:

\frac{df}{dx} = \frac{d}{dx}(x^2 + 1)^{5}

⇒ \frac{df}{dx} = 5(x^2 + 1)^4\frac{d}{dx}( x^2 + 1)

⇒ \frac{df}{dx} = 5(x^2 + 1)^4(\frac{d}{dx}( x^2))

⇒\frac{df}{dx} = 5(x^2 + 1)^4(2x)

⇒\frac{df}{dx} = 10x(x^2 + 1)^4

**Examples 5: Find the derivative of the function, f(x) = e(2x + 5).

**Solution:

f(x) = e(2x + 5)

\frac{df}{dx} = \frac{d}{dx}(e^{(2x +5)})

⇒\frac{df}{dx} = \frac{d}{dx}(e^{(2x +5)})

⇒\frac{df}{dx} = e^{(2x +5)}\frac{d}{dx}(2x + 5)

⇒\frac{df}{dx} = e^{(2x +5)}2

⇒\frac{df}{dx} = 2e^{(2x +5)}

**Examples 4: Find the derivative of the function f(x) = sin(tan x + 5).

**Solution:

f(x) = sin(tan x + 5)

**⇒\frac{d}{dx}f(x) = \frac{d}{dx} sin(tan(x) + 5)

**⇒\frac{d}{dx}f(x) = cos(tan(x) + 5)\frac{d}{dx}(tan(x) + 5)

**⇒\frac{d}{dx}f(x) = cos(tan(x) + 5)\frac{d}{dx}(tan(x))

**⇒\frac{d}{dx}f(x) = cos(tan(x) + 5)sec^2(x)

**Examples 2: Find the derivative for the function f(x) = (x6 + x2 + 1)10

**Solution:

\frac{df}{dx} = \frac{d}{dx}(x^6 + x^2 + 1)^{10}

⇒ \frac{df}{dx} = 10(x^6 + x^2 + 1)^9\frac{d}{dx}(x^6 + x^2 + 1)

⇒ \frac{df}{dx} = 10(x^6 + x^2 + 1)^9(\frac{d}{dx}x^6 + \frac{d}{dx}x^2)

⇒\frac{df}{dx} = 10(x^6 + x^2 + 1)^9(6x^5 +2x)

Practice Questions on Derivative of Composite Functions

**Q1: Differentiate sin(log x)

**Q2: Find the derivative of e(sin x + cos x)

**Q3: Differentiate cos(x2 + 2x)

**Q4: Differentiate tan x2 + sec2x