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

**Solution:**

Let f(x) = sin (ax + b),

u(x) = ax + b and v(t)

= sint

Then, (vou)(x) = v(u(x))

= v(ax + b)

= sin(ax + b) = f(x)

Here, f is a composite function of two functions u and v.

Put, t = u(x) = ax + b

dv / dt = d / dt (sin t)

= cost = cos(ax + b)

⇒ dt / dx = d / dx (ax + b)

= d / dx (ax) + d / dx (b)

= a + 0 = a

Hence, by chain rule, we get

df/dx = (dv / dt) × (dt / dx)

= cos(ax + b).a = a cos (ax + b)

Alternate method:

d/dx [sin (ax + b)] = cos (ax + b) .d / dx (ax + b)

= cos(ax + b).[d / dx (ax) + d / dx (b)]

= cos (ax + b).(a + 0)

= a cos(ax + b)

NCERT Solutions Class 12 Maths - Chapter 5 Exercise 5.2 Question 3

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

**Summary:**

By using chain rule we have the derivative of sin(ax+b) is a cos(ax + b)