Rewriting an Exam Question

When answering exam questions some weak students normally rewrite the question in the Answers booklet before providing the solution.why do they rewrite it?it does not add any value. Moreover it is a waste of precious time. No matter how many times you tell them they always repeat.Some explain this behaviour by saying the tutor may confuse the solution and its respective item.

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}