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:

- x
^{3}– 2x^{2}+ 2 - y = 5x
^{4}+ 10x^{2}+ 2 - y = (5x + 2)(2x
^{2}+ 2) - y = x
^{-3}(5 +2x) - S = with respect to r (written as wrt r)

## Now let us solve these examples of differentiation:

**Example 1**: x^{3} – 2x^{2} + 2

=

= 3x^{2} – 2 x 2x + 0

= 3x^{2} – 4x

**Example 2**: y = 5x^{4} + 10x^{2} + 2

=

= 4 x 5x^{3} + 2 x 10x + 0

= 20x^{3} + 20x

**Example 3**: y = (5x + 2)(2x^{2} + 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) = (2x^{2} + 3)f'(x) = 5

g'(x) = 4x

= f'(x) g(x) + g'(x) f(x)

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

= 10x^{2} + 15 + 20x^{2} + 8x

= 30x^{2} + 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) = 2f(x)g(x)

= f'(x)g(x) + g'(x)f(x)

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

=

=

=

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

y = =

y’ =

=

=

= (divide numerator and denominator by x^{2})

=

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

**Example 5**: S =

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

= +

= Remember 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