# Chain Rule

## Related calculator: Online Derivative Calculator with Steps

Now let's see how to differentiate composite functions.

Suppose that we are given function h(x)=f(g(x)). Remembering that g'(x) is rate of change of g(x) with respect to x and f'(g(x)) is rate of change of f with respect to g(x) then it is reasonable to suggest that rate of change of f with respect to x is product of f'(g(x)) and g'(x).

Indeed, if g changes twice as fast as x and f changes three times as fast as g, then f changes six times as fast as x.

Chain Rule. If f and g are both differentiable and h=f@g is the composite function defined by h(x)=f(g(x)), then h is differentiable and h'(x)=f'(g(x))g'(x).

Proof. Recall that if y=f(x) and x changes from a to a+Delta x then increment of y is Delta y=f(a+Delta x)-f(a) . According to the definition of derivative lim_(Delta x->0)(Delta y)/(Delta x)=f'(a).

So if we denote by epsi the difference between the difference quotient and the derivative, we obtain lim_(Delta x->0)epsi=lim_(Delta x->0)((Delta y)/(Delta x)-f'(a))=f'(a)-f'(a)=0.

But epsi=(Delta y)/(Delta x)-f'(a) or Delta y=f'(a)Delta x+epsi Delta x.

Thus, for any differentiable function f, Delta y=f'(a) Delta x+epsi Delta x where epsi->0 as Delta x->0.

Now, suppose u=g(x) is differentiable at a and y=f(u) is differentiable at b=g(a). If Delta x is an increment in x and Delta u and Delta y are the corresponding increments in u and y then

Delta u=g'(a)Delta x+epsi_1 Delta x=(g'(a)+epsi_1)Delta x where epsi_1->0 as Delta x->0

Delta y=f'(b)Delta u+epsi_2 Delta u=(f'(b)+epsi_2)Delta u where epsi_2->0 as Delta u->0

Now substitute expression for Delta u in the last equation:

Delta y=(f'(b)+epsi_2)(g'(a)+epsi_1)Delta x

Or

(Delta y)/(Delta x)=(f'(b)+epsi_2)(g'(a)+epsi_1)

As Delta x->0 thenDelta u->0. So, both epsi_1->0 and epsi_2->0 as Delta x->0 .

Therefore, (dy)/(dx)=lim_(Delta x->0)((f'(b)+epsi_2)(g'(a)+epsi_1))=f'(b)g'(a)=f'(g(a))g'(a).

In Leibniz notation, if y=f(u) and u=g(x) are both differentiable then (dy)/(dx)=(dy)/(du)(du)/(dx).

In Leibniz notation it is especially easy to remember chain rule because if (dy)/(du) and (du)/(dx) were quotients, then we could cancel du. Remember, however, that du has not been defined and (du)/(dx) should not be thought of as an actual quotient.

Example 1. Find derivative of h(x)=sqrt(x^2+1)

Here f(u)=sqrt(u) and g(x)=x^2+1 and h(x)=f(g(x)), therefore

f'(u)=(sqrt(u))'=1/(2sqrt(u)) and g'(x)=(x^2+1)'=2x.

So, h'(x)=f'(g(x))g'(x)=f'(sqrt(x^2+1))2x=1/(2sqrt(x^2+1))2x=x/(sqrt(x^2+1)).

In using the Chain Rule we work from the outside to the inside. We differentiate the outer function [at the inner function g(x)] and then we
multiply by the derivative of the inner function.

Example 2. Differentiate y=cos(x^3) and y=(cos(x))^3 .

If y=cos(x^3) then outer function is cosine and inner is cubic function, so y'=-sin(x^3)*(x^3)'=-3x^2sin(x^3) .

If y=(cos(x))^3 then outer function is cubic and inner is cosine, so y'=3(cos(x))^2*(cos(x))'=-3(cos(x))^2sin(x).

Example 3. Differentiate y=(x^2+1)^7

y'=7(x^2+1)^6*(x^2+1)'=7(x^2+1)^6*2x=14x(x^2+1)^6.

Example 4. Differentiate y=((2t+3)/(t-5))^8

Here we use chain rule and quotient rule.

y'=8((2t+3)/(t-5))^(8-1)*((2t+3)/(t-5))'=8((2t+3)/(t-5))^7 ((2t+3)'(t-5)-(2t+3)(t-5)')/(t-5)^2=

=8((2t+3)/(t-5))^7 (2(t-5)-(2t+3))/(t-5)^2=8((2t+3)/(t-5))^7 (-13)/(t-5)^2=-104 ((2t+3)^7)/((t-5)^9).

Example 5. Find derivative of f(x)=(3x^2+4x+1)^5(e^x+sin(x))^2.

We need to use product rule together with chain rule.

f'(x)=((3x^2+4x+1)^5)'(e^x+sin(x))^2+(3x^2+4x+1)^5((e^x+sin(x))^2)'=

=5(3x^2+4x+1)^4*(3x^2+4x+1)'(e^x+sin(x))^2+(3x^2+4x+1)^5 2(e^x+sin(x))(e^x+sin(x))'=

=5(3x^2+4x+1)^4*(6x+4)(e^x+sin(x))^2+(3x^2+4x+1)^5 2(e^x+sin(x))(e^x+cos(x))=

=2(3x^2+4x+1)^4(e^x+sin(x))(5(3x+2)(e^x+sin(x))+(3x^2+4x+1)(e^x+cos(x))).

Now let's see how to use chain rule more than once.

Example 6. Differentiate f(t)=e^(cos(2t)).

We apply chain rule twice.

f'(t)=(e^(cos(2t)))'=e^(cos(2t))*(cos(2t))'=e^(cos(2t))*(-sin(2t))*(2t)'=-2sin(2t)e^(cos(2t)).

Example 7. Differentiate f(x)=cos(sin(tan(x)))

Here we again apply chain rule twice.

f'(x)=(cos(sin(tan(x))))'=-sin(sin(tan(x)))*(sin(tan(x)))'=

=-sin(sin(tan(x)))*(cos(tan(x)))*(tan(x))'=-sec^2(x)cos(tan(x))sin(sin(tan(x))).

In general we can apply chain rule even more than two times. We need to use it as many times as we need.