PRODUCT RULE OF DERIVATIVE

Let u and v be two functions of x.

If u and v are multiplied, then the following is the rule to find derivative of the product of u and v with respect to x.

(uv)' = uv' + u'v

The above is called the product rule of derivative.

Example 1 :

Find dy/dx, if y = (x³ + x²)(x + 5).

Solution :

y = (x³ + x²)(x + 5)

Let u = x³ + x² and v = x + 5.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = x³ + x² and v = x + 5.

dy/dx = (x³ + x²)(x + 5)' + y + (x³ + x²)'(x + 5)

= (x³ + x²)(1 + 0) + (3x^{3 - 1} + 2x^{2 - 1})(x + 5)

= (x³ + x²)(1) + (3x² + 2x)(x + 5)

= x³ + x² + (3x³ + 15x² + 2x² + 10x)

= x³ + x² + (3x³ + 17x² + 10x)

= x³ + x² + 3x³ + 17x² + 10x

= 4x³ + 18x² + 10x

Example 2 :

Find dy/dx, if y = (-2x⁴ - 3)(-2x² + 1).

Solution :

y = (-2x⁴ - 3)(-2x² + 1)

Let u = -2x⁴ - 3 and v = -2x² + 1.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = -2x⁴ - 3 and v = -2x² + 1.

dy/dx = (-2x⁴ - 3)(-2x² + 1)' + (-2x⁴ - 3)'(-2x² + 1)

= (-2x⁴ - 3)(-2 ⋅ 2x^{2 - 1} + 0) + (-2 ⋅ 4x^{4 - 1} - 0)(-2x² + 1)

= (-2x⁴ - 3)(-4x) + (-8x³)(-2x² + 1)

= (-2x⁴ - 3)(-4x) - 8x³(-2x² + 1)

= 8x⁵ + 12x + 16x⁵ - 8x³

= 24x⁵ - 8x³ + 12x

Example 3 :

Find dy/dx, if y = e^xlnx.

Solution :

y = e^xlnx

Let u = e^x and v = lnx.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = e^x and v = lnx.

dy/dx = e^x(lnx)' + (e^x)'lnx

Example 4 :

Find dy/dx, if y = 2^xx⁵.

Solution :

y = 2^xx⁵

Let u = 2^x and v = x⁵.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = 2^x and v = x⁵.

dy/dx = 2^x(x⁵)' + (2^x)'x⁵

= 2^x(5x^{5 - 1}) + (2^xln2)x⁵

= 2^x5x⁴ + x⁵2^xln2

= x⁴2^x(5 + xln2)

Example 5 :

Find dy/dx, if y = 2^xlnx.

Solution :

y = 2^xlnx

Let u = 2^x and v = lnx.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = 2^x and v = lnx.

dy/dx = 2^x(lnx)' + (2^x)'lnx

Example 6 :

Find dy/dx, if y = x³sinx.

Solution :

y = x³sinx

Let u = x³ and v = sinx.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = x³ and v = sinx.

dy/dx = x³(sinx)' + (x³)'sinx

= x³(cosx) + (3x^{3 - 1})sinx

= x³cosx + 3x²sinx

= x²(xcosx + 3sinx)

Example 7 :

Find dy/dx, if y = e^2xtanx.

Solution :

y = e^2xtanx

Let u = e^2x and v = tanx.

y = uv

dy/dx = (uv)'

Using product rule of derivative,

dy/dx = uv' + u'v

Substitute u = e^2x and v = tanx.

dy/dx = e^2x(tanx)' + (e^2x)'tanx

= e^2x(sec²x) + (2e^2x)tanx

= e^2x(sec²x + 2tanx)

Example 8 :

Find dy/dx, if y = 3sin³xcosx.

Solution :

y = 3sin³xcosx

Let u = sin³x and v = cosx.

y = 3uv

dy/dx = 3(uv)'

Using product rule of derivative,

dy/dx = 3(uv' + u'v)

Substitute u = sin³x and v = cosx.

dy/dx = 3[sinx³x(cosx)' + (sin³x)'cosx]

= 3[sin³x(-sinx) + (3sin^3-1xcosx)cosx]

= 3[-sin³xsinx + (3sin²xcosx)cosx]

= 3[-sin³xsinx + 3sin²xcos²x]

= -3sin³xsinx + 9sin²xcos²x

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