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Examples of Differentiation

Here are a few examples of differentiation for various types of equations – obtained by by applying the rules (product rule, quotient rule, etc) relevant to each type of equation.

Find the derivatives for the following equations:

  1. x3 – 2x2 + 2
  2. y = 5x4 + 10x2 + 2
  3. y = (5x + 2)(2x2 + 2)
  4. y = x-3(5 +2x)
  5. S =  2pi r^2 + 2pi rh with respect to r (written as wrt r)

 

Now let us solve these examples of differentiation:

Example 1:  x3 – 2x2 + 2

{d}/{dx}(x^3~-~2x^2~+~2)

=  {d}/{dx}x^3~+~{d}/{dx}(-2x^2)~+~{d}/{dx}2

= 3x2 – 2 x 2x + 0

= 3x2 – 4x

 

Example 2: y = 5x4 + 10x2 + 2

{d}/{dx}(5x^4~+~10x^2~+~2)

=  {d}/{dx}5x^4~+~{d}/{dx}10x^2~+~{d}/{dx}2

= 4 x 5x3 + 2 x 10x + 0

= 20x3 + 20x

 

Example 3:  y = (5x + 2)(2x2 + 3)

We can express this function as a product of two equations, and apply the product rule for differentiation.

f(x) = (5x + 2)                g(x) = (2x2 + 3)f'(x) = 5

g'(x) = 4x{dy}/{dx}~=~{d}/{dx}f(x)~g(x)

=  f'(x) g(x)  +  g'(x) f(x)

= 5(2x2 + 3) + 4x(5x + 2)

= 10x2 + 15 + 20x2 + 8x

= 30x2 + 8x + 15

 

Example 4: y = x-3(5 +2x)

Again we can express this function as a product of two equations, and apply the product rule for differentiation.

f(x) = x-3                         g(x) = (5 + 2x)f'(x) = -3x-4 

g'(x) = 2{dy}/{dx}~=~{d}/{dx}f(x)g(x)

=  f'(x)g(x) + g'(x)f(x)

= (-3x-4)(5 + 2x) + 2x-3

= {-3}/{x^4}(5~+~2x)~+~{{2}/{x^3}}

= {{-3}(5~+~2x)~+~{2x}}/{x^4}

= {-4x~-~15}/{x^4}

 

We could have solved the above equation using the Quotient Rule also, as illustrated below –

y  =  {(5~+~2x)}/{x^3}  =  {f(x)}/{g(x)}

y’  =  {{{f}prime{(x)}}{g(x)}~-~{{f(x)}{g}prime{(x)}}}/{(g(x))^2}

=  {{x^3}{(2)}~-~{(5~+~2x)}{(3x^2)}}/{(x^3)^2}

=  {2{x^3}~-~{3(5~+~2x)}{(x^2)}}/{(x^6)}

=  {2{x}~-~{3(5~+~2x)}}/{x^4}     (divide numerator and denominator by x2)

= {-4x~-~15}/{x^4}

This is the same answer we got above by using the product rule.

 

Example 5: S =  2pi r^2 +~2pi rh

Differentiating with respect to r (written as wrt r) –

{ds}/{dr}  =  {d/{dr}}(2pi r^2) + {d/{dr}}(2pi rh)

=  4pi r~+~2 pi h Remember pi and h are constants here, only r is the variable that can be differentiated.

 

Learn more about first principles of derivatives, and finding derivatives using formula