Basic techniques of differentiation

Techniques of Differentiating

The following are basic techniques of differentiating. 

(a) Power rule

So far we have seen that given; y=f(x)=kx^{n}

Then the gradient function is; dy/dx=knx^{n-1}

Example 1: Differentiate the function y=3x^{5} with respect to x

solution: dy/dx=15x^{4}

Example 2: Find the derivative of the function y=2x^{3}-6x+3

solution : dy/dx=6x^{2}-6

Example 3: Differentiate the function y= [6x^{3}-7x]/[x^{2}]

solution: we need to first simplify the function by rewriting it as;

y=6x^{3}/x^{2}-7x/x^{2}=6x-7x^{-1}

Hence we have;

          dy/dx =6+7x^{-2}

(b) Product rule

Let y=f(x)=uv where u and v are both differentiable functions of x, then the gradient function is given as;

dy/dx= u dv/dx+v du/dx

Alternatively;

 (uv)^/=uv^/ +vu^/

This rule is appropriate when it comes to integrating product functions.

Example 1: Given y=(3x^{4}+7x)(5x^{7}-3x^{2}+9x) determine the derivative.

Solution: by definition dy/dx=u dv/dx+ v du/dx

we let,

u=3x^{4}+7x then du/dx=12x^{3}+7  and we let v=5x^{7}-3x^{2}+9x  then dv/dx =35x^{6}-6x+9

Therefore; 

dy/dx=u dv/dx+ v du/dx=(3x^{4}+7x)(35x^{6}-6x+9)+(5x^{7}-3x^{2}+9x)(12x^{3}+7)

Example 2: Given y=3x^{5}sinx , determine the gradient function.

Solution: by definition dy/dx =u dv/dx+v du/dx

We let, u=3x^{5} then du/dx =15x^{4}

Also let, v = sinx    then  dv/dx =cosx

Therefore; 

dy/dx=u dv/dx+v du/dx =3x^{5}cos x+15x^{4}sin x=3x^{4}(xcos x+5sin x)

Example 3: Let y=3x^{2}e^{x}, find dy/dx

Solution: We let u=3x^{2} then du/dx=6x

Also v=e^{x} then dv/dx=e^{x}

Hence;

dy/dx=(3x^{2})(e^{x})+(6x)(e^{x})=3x^{2}e^{x}+6xe^{x}

(c)  Quotient rule

Let y=f(x)= u/v  where u and  v are both differentiable functions of  x, then the gradient function is given as;

dy/dx=[v du/dv –u dv/dx]/v^{2}

Example 1:  Given y= 4x^{3}/[3x^{2}-4x]  find dy/dx;

Solution: Let u=4x^{3} then du/dx =12x^{2} and v =3x^{2}-4x then dv/dx = 6x-4

dy/dx=[12x^{2}(3x^{2}-4x)-4x^{3}(6x-4)}]/[(3x^{2}-4x)^{2}]

=[36x^{4}-48x^{3}-24x^{4}+16x^{3}]/[{(3x^{2}-4x)^{2}}]

=[12x^{4}-32x^{3}]/[(3x^{2}-4x)^{2}]

=[4x^{3}(3x-8)]/[(3x^{2}-4x)^{2}]

Example 2: Find the derivative of the function;

y= 2x/sin x

Solution: Let;

u = 2x then du/dx  =2

Let; v= sin x then dv/dx=cos x

Hence;

dy/dx=[2sin x -2xcos x]/sin^2x=2[sin x – x cos x]/ sin^{2}x

(d) Chain rule

Let y be a differentiable function of  u  i.e.  y=f(u)  and that u  is a differentiable function of x i.e. u=g(x), then y is a differentiable function of x i.e. y=f(g(x)), and;

dy/dx=(dy/du )(du/dx)

Example 1: Suppose y=(3x^{5}+7x^{4})^{4}, find dy/dx

Solution:

Let u=3x^{5}+7x^{4} then  du/dx=15x^{4}+28x^3

Also y=u^{4} then  dy/du=4u^{3}

By definition

dy/dx=(dy/du)(du/dx)=(4u^{3})(15x^{4}+28x^3)

=4(15x^{4}+28x^3)(3x^{5}+7x^{4})^{3}

=4x^{3}(15x+28)(3x^{5}+7x^{4})^{3}