Question

Differentiate the function with respect to x.
$sin(ax + b)$

Answer 1

Hint: Differentiation is a method to find the slope of functions on a graph. In complex functions, the chain rule is used for differentiation. Mathematically, it is given by-
$\dfrac{{d{\text{y}}}}{{d{\text{x}}}} = \dfrac{{d{\text{y}}}}{{d{\text{t}}}} \times \dfrac{{d{\text{t}}}}{{d{\text{x}}}}...\left( 1 \right)$

Complete step-by-step answer:
Let the function be-
${\text{y}} = \sin \left( {ax + {\text{b}}} \right)$...(2)
We can clearly see that $(ax + b)$ is a function within the sine function itself. So let us assume that-
$t = ax + b$ …(3)

So, we can differentiate equation (2) with respect to t as-
$\begin{aligned} &{\text{y}} = \sin \left( {ax + {\text{b}}} \right) = sint \\\ &\dfrac{{dy}}{{dt}} = cost \\\ \end{aligned}$

Also differentiating equation (3) with respect to x as-
$\begin{aligned} &{\text{t}} = ax + {\text{b}} \\\ & \dfrac{{dt}}{{dx}} = {\text{a}} \\\ \end{aligned}$

Applying these values in equation (1) we can get-
$\begin{aligned} &\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{dt}} \times \dfrac{{dt}}{{dx}} \\\ &\dfrac{{dy}}{{dx}} = cost \times {\text{a}} \\\ &\dfrac{{dy}}{{dx}} = acost \\\ \end{aligned}$

We can substitute the value of t from equation (3) as-
$\dfrac{{dy}}{{dx}} = acos\left( {ax + {\text{b}}} \right)$
This is the required answer.

Note: In such types of questions, we should not differentiate directly as it will always result in a wrong answer. The chain rule can be applied for any number of functions. The general form of the chain rule is-
$\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{d{t_1}}} \times \dfrac{{d{t_1}}}{{d{t_2}}} \times \dfrac{{d{t_2}}}{{d{t_3}}} \times ... \times \dfrac{{d{t_{{\text{n}} - 1}}}}{{d{t_{\text{n}}}}} \times \dfrac{{d{t_{\text{n}}}}}{{dx}}$