Contact Submit FAQ About

Calculus - Differentiation - Product Rule

  a - answer    s - solution    v - video

Question 1


Use the product rule: when $y=u\times v$, $\frac{\mathrm{d}y}{\mathrm{d}x}=v\frac{\mathrm{d}u}{\mathrm{d}x}+u\frac{\mathrm{d}v}{\mathrm{d}x}$ to differentiate the following:

a) $y=2x(3x+1)^3$  a s v
b) $y=5x(2x+3)^4$  a s v
c) $y=x(2x+5)^2$  a s v
d) $y=3x(1-x)^3$  a s v
e) $y=2x{(3x+2)}^{-1}$  a s v
f) $y=4x{(2-7x)}^{-6}$  a s v


Question 2


Use the product rule to differentiate the following:

a) $y={x^2}{(3x+5)}^5$  a s v
b) $y={x^3}{(5-x)}^6$  a s v
c) $y={2x^5}{(8x+3)}^{\frac{3}{2}}$  a s v
d) $y={6x^{-1}}{(2-5x)}^{3}$  a s v
e) $y={3x^{\frac43}}{(2-x)}^{-1}$  a s v


Question 3


Use product rule to differentiate:

a) $y=2x{(x^2-1)}^3$  a s v
b) $y=-4x{(x^3+8)}^5$  a s v
c) $y=x^2{(x^3+2)}^4$  a s v
d) $y=5x^2{(2x^4-8)}^{10}$  a s v
e) $y=10x^{\frac12}{(4x+6)}^{-1}$  a s v
f) $y=-8x^{2}{(4-5x)}^{-\frac54}$  a s v
g) $y={(5-2x^2)}^3\frac12x^{-1}$  a s v


Question 4


Use product rule to differentiate:

a) $y=(x+1)(x+2)$  a s v
b) $y={(x+1)}^2(x+2)$  a s v
c) $y={(x+1)}^2{(x+2)}^2$  a s v
d) $y={(2x+5)}^2{(x+5)}^3$  a s v
e) $y={(3-2x)}^3{(5x+7)}^5$  a s v
f) $y={(x^2+2)}{(x-3)}$  a s v
g) $y={(3x^2-2)}^3{(8x-4)}^4$  a s v
h) $y={(4x^3+2)}^3{(8x^2-3x)}^4$  a s v
i) $y={(3x^{-1}+2)}^3{(4x^2+3)}^{-1}$  a s v
j) $y=\sqrt{(2x+3)}{(3x+1)}$  a s v