Product Rule

The product rule is a fundamental differentiation technique used when you need to differentiate the product of two functions. It allows you to find the derivative of a product without having to multiply the functions first, which can simplify the differentiation process, especially when dealing with complex functions.

If a function can be written as :

y=uv

Then the first derivative is given by :

\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}

Example

infoNote

Differentiate $x^2 \cdot \sin{(x)}$

First, identify the product that is occurring :

u=x^2

v=\sin{(x)}

Differentiate each function :

u' = 2x

v' = \cos(x)

Apply product rule :

\frac{dy}{dx} = u(x) \cdot v'(x) + v(x) \cdot u'(x)

\frac{dy}{dx} = x^2 \cdot \cos(x) + \sin(x) \cdot 2x

= x^2 \cos(x) + 2x \sin(x)

Example

infoNote

Differentiate $y=e^x \cdot \ln{(x)}$

Identify the product :

u=e^x

v=\ln{(x)}

Differentiate each function :

u' = e^x

v' = \frac{1}{x}

Apply product rule :

\frac{dy}{dx} = u(x) \cdot v'(x) + v(x) \cdot u'(x)

\frac{dy}{dx} = e^x \cdot \frac{1}{x} + \ln(x) \cdot e^x

\frac{e^x}{x} + e^x \ln(x)

Example

infoNote

Differentiate $y = x^2 \sqrt{3x+1}$

Identify the product :

u=x^2

v=\sqrt{3x+1}

Differentiate each function :

$u'$

u' = 2x

$v'$ (using chain rule)

infoNote

v' = \frac{1}{2} (3x+1)^{-\frac{1}{2}} \times 3 = \frac{3}{2} (3x+1)^{-\frac{1}{2}}

Apply product rule :

\frac{dy}{dx} = u'v + uv'

= (3x+1)^{\frac{1}{2}} (2x) + x^2 \left( \frac{3}{2} (3x+1)^{-\frac{1}{2}} \right)

= 2x (3x+1)^{\frac{1}{2}} + \frac{3x^2}{2} (3x+1)^{-\frac{1}{2}}

= x (3x+1)^{-\frac{1}{2}} \left[ 2(3x+1) + \frac{3x}{2} \right]

= x (3x+1)^{-\frac{1}{2}} \left[ 6x + 2 + \frac{3x}{2} \right]

= x (3x+1)^{-\frac{1}{2}} \left[ \frac{15x+2}{2} \right]

= \frac {1}{2}x (3x+1)^{-\frac{1}{2}} \left( 15x+4\right)

= \frac {15x^2+4x}{2 \sqrt{3x+1}}

:::

Product Rule Simplified Revision Notes for Leaving Cert Mathematics

Product Rule

Differentiate $x^2 \cdot \sin{(x)}$

Differentiate $y=e^x \cdot \ln{(x)}$

Differentiate $y = x^2 \sqrt{3x+1}$

500K+ Students Use These Powerful Tools to Master Product Rule For their Leaving Cert Exams.

Other Revision Notes related to Product Rule you should explore

Product Rule Simplified Revision Notes for Leaving Cert Mathematics

Product Rule

Differentiate x2⋅sin⁡(x)x^2 \cdot \sin{(x)}x2⋅sin(x)

Differentiate y=ex⋅ln⁡(x)y=e^x \cdot \ln{(x)}y=ex⋅ln(x)

Differentiate y=x23x+1y = x^2 \sqrt{3x+1}y=x23x+1​

500K+ Students Use These Powerful Tools to Master Product Rule For their Leaving Cert Exams.

Other Revision Notes related to Product Rule you should explore

Differentiate $x^2 \cdot \sin{(x)}$

Differentiate $y=e^x \cdot \ln{(x)}$

Differentiate $y = x^2 \sqrt{3x+1}$