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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′] / h2
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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