Derivatives — MATHS STD 11 Science — Question

Let $\text{u}=(\text{ax}+\text{b})^\text{n},\text{v}=(\text{cx}+\text{d})^\text{m}$

Then, $\text{u}'=\text{na}(\text{ax}+\text{b})^{\text{n}-1},\text{v}'=\text{nc}(\text{cx}+\text{d})^{\text{m}-1}$

Using the product rule:

$\frac{\text{d}}{\text{dx}}(\text{uv})=\text{uv}'+\text{vu}'$

$\frac{\text{d}}{\text{dx}}[(\text{ax}+\text{b})^\text{n}(\text{cx}+\text{d})^\text{m}]=(\text{ax}+\text{b})^\text{n}\times\text{mc}(\text{cx}+\text{d})^{\text{m}-1}+(\text{cx}+\text{d})^\text{m}\times\text{na}(\text{ax}+\text{b})^{\text{n}-1}$

$\text{mc}(\text{ax}+\text{b})^\text{n}(\text{cx}+\text{d})^{\text{m}-1}+\text{na}(\text{cx}+\text{d})^\text{m}(\text{ax}+\text{b})^{\text{n}-1}$