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Miscellaneous Exercise · Q20

Q.Find the derivative of a+bsin⁡xc+dcos⁡x\dfrac{a + b\sin x}{c + d\cos x}.

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The derivative of a quotient of two functions is found using the quotient rule: ddx(uv)=u′v−uv′v2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}. For f(x)=a+bsin⁡xc+dcos⁡xf(x) = \frac{a + b\sin x}{c + d\cos x}, the derivative is bcos⁡x(c+dcos⁡x)+dsin⁡x(a+bsin⁡x)(c+dcos⁡x)2\frac{b\cos x (c + d\cos x) + d\sin x (a + b\sin x)}{(c + d\cos x)^2}.

The quotient rule is the natural choice here because we have one function divided by another. Instead of memorising a formula blindly, think of it this way: when you have a fraction, the rate of change of the whole depends on how the numerator and denominator change relative to each other. The quotient rule captures that by taking the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

Let’s apply it step by step.

  1. Identify the numerator and denominator

    Let u=a+bsin⁡xu = a + b\sin x and v=c+dcos⁡xv = c + d\cos x.

    The function is f(x)=uvf(x) = \frac{u}{v}.

  2. Find the derivatives of uu and vv

    • u′=ddx(a)+ddx(bsin⁡x)=0+bcos⁡x=bcos⁡xu' = \frac{d}{dx}(a) + \frac{d}{dx}(b\sin x) = 0 + b\cos x = b\cos x
    • v′=ddx(c)+ddx(dcos⁡x)=0−dsin⁡x=−dsin⁡xv' = \frac{d}{dx}(c) + \frac{d}{dx}(d\cos x) = 0 - d\sin x = -d\sin x
    Watch out

    A common mistake is forgetting the sign when differentiating cos⁡x\cos x. Remember: ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x, so dcos⁡xd\cos x gives −dsin⁡x-d\sin x.

  3. Apply the quotient rule

    The quotient rule states:

f′(x)=u′v−uv′v2f'(x) = \frac{u'v - uv'}{v^2}

Substitute uu, vv, u′u', and v′v':

f′(x)=(bcos⁡x)(c+dcos⁡x)−(a+bsin⁡x)(−dsin⁡x)(c+dcos⁡x)2f'(x) = \frac{(b\cos x)(c + d\cos x) - (a + b\sin x)(-d\sin x)}{(c + d\cos x)^2}

  1. Simplify the numerator Notice the minus sign in front of the second term: =bcos⁡x(c+dcos⁡x)+dsin⁡x(a+bsin⁡x)(c+dcos⁡x)2= \frac{b\cos x (c + d\cos x) + d\sin x (a + b\sin x)}{(c + d\cos x)^2} …

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