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NCERT Exemplar · Q37

Q.Differentiate with respect to xx: a+bsin⁡xc+dcos⁡x\dfrac{a + b\sin x}{c + d\cos x}.

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Apply the Quotient Rule to differentiate a fraction where both numerator and denominator involve trigonometric functions; the derivative is b(c+dcos⁡x)cos⁡x+d(a+bsin⁡x)sin⁡x(c+dcos⁡x)2\boxed{\dfrac{b(c+d\cos x)\cos x + d(a+b\sin x)\sin x}{(c+d\cos x)^2}}.

When you have a function written as one expression divided by another, the Quotient Rule is your tool. The idea is simple: the rate of change of a fraction depends on how fast the top is changing relative to the bottom, adjusted for the bottom's own rate of change. If uu and vv are both functions of xx, then

ddx(uv)=v⋅u′−u⋅v′v2.\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \cdot u' - u \cdot v'}{v^2}.

The numerator of the result captures the interplay: the denominator's value times the numerator's derivative, minus the numerator's value times the denominator's derivative. The denominator squared ensures the units work out and reflects how a faster-changing denominator amplifies the overall rate of change.

Here, u=a+bsin⁡xu = a + b\sin x and v=c+dcos⁡xv = c + d\cos x. Both a,b,c,da, b, c, d are constants.


Step-by-step differentiation:

  1. Identify the pieces.

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

  2. Differentiate the numerator.

    The derivative of a constant is zero, and ddx(sin⁡x)=cos⁡x\frac{d}{dx}(\sin x) = \cos x, so

u′(x)=0+bcos⁡x=bcos⁡x.u'(x) = 0 + b\cos x = b\cos x.

  1. Differentiate the denominator. Similarly, ddx(cos⁡x)=−sin⁡x\frac{d}{dx}(\cos x) = -\sin x, so

v′(x)=0+d(−sin⁡x)=−dsin⁡x.v'(x) = 0 + d(-\sin x) = -d\sin x.

  1. Apply the Quotient Rule. Substitute into v⋅u′−u⋅v′v2\frac{v \cdot u' - u \cdot v'}{v^2}:

ddx(a+bsin⁡xc+dcos⁡x)=(c+dcos⁡x)(bcos⁡x)−(a+bsin⁡x)(−dsin⁡x)(c+dcos⁡x)2.\frac{d}{dx}\left(\frac{a+b\sin x}{c+d\cos x}\right) = \frac{(c+d\cos x)(b\cos x) - (a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}.

  1. Simplify the numerator. Distribute in the first term:

(c+dcos⁡x)(bcos⁡x)=bccos⁡x+bdcos⁡2x.(c+d\cos x)(b\cos x) = bc\cos x + bd\cos^2 x.

In the second term, the double negative becomes a plus:

−(a+bsin⁡x)(−dsin⁡x)=d(a+bsin⁡x)sin⁡x=adsin⁡x+bdsin⁡2x.-(a+b\sin x)(-d\sin x) = d(a+b\sin x)\sin x = ad\sin x + bd\sin^2 x.

Combine:

bccos⁡x+bdcos⁡2x+adsin⁡x+bdsin⁡2x.bc\cos x + bd\cos^2 x + ad\sin x + bd\sin^2 x. …

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