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# Differentiate the function with respect to x. sin(ax+b)/cos(cx+d)

**Solution:**

Given, f(x) = sin (ax + b) cos(cx + d),

where g(x) = sin(ax + b) and h(x) = cos(cx + d)

⇒ f = [g′h−g h′] / h^{2}

Consider g(x) = sin (ax + b)

Let u(x) = ax + b, v(t) = sin t

Then (vou) (x) = v(u(x)) = v (ax + b)

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

Therefore,

g is a composite function of two functions, u* *and v*.*

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

dv/dt = d/dt (sin t)

= cos t = cos (ax + b)

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

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

= a + 0 = a

Thus, by chain rule, we get

g′ = dg/dx = dv/dt .dt/dx

= cos (ax + b).a

= a cos (ax + b)

Consider h(x) = cos(cx + d)

Let p(x) = cx + d,

q(y) = cos y

Then, (q o p)(x) = q(p(x))

= q (cx + d)

= cos (cx + d) = h(x)

⇒ h is a composite function of two functions, p* *and* q.*

Put, y = p(x) = cx+d

⇒ dq / dy = d/dy (cosy)

= −sin y = −sin (cx + d)

dy/dx = d/dx (cx + d)

= d/dx (cx) + d/dx (d) = c

Using chain rule, we get

h′ = dh/dx

= dq/dy .dy/dx

= −sin (cx + d) × c

= −c sin (cx + d)

Therefore,

f′ = a cos (ax + b).cos (cx + d) − sin(ax + b) {− c sin (cx + d)} / [cos (cx + d)]^{2}

=[ a cos (ax + b) cos (cx + d) + c sin (ax + b). sin (cx + d)] / [cos (cx + d) × cos (cx + d)]

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

## Differentiate the function with respect to x. sin(ax+b)/cos(cx+d)

**Summary:**

The derivative of sin(ax+b) / cos(cx+d) is [a cos (ax + b) cos (cx + d) + c sin (ax + b). sin (cx + d)] / [cos (cx + d) × cos (cx + d)]

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