Test Index

NCERT Class XI Mathematics - Limits and Derivatives - Solutions

© examsnet.com
Question : 63 of 72
Marks: +1, -0
a+bsin⁡xc+dcos⁡x\frac{a+b\sin x}{c+d\cos x}
Solution:  
Let f (x) = a+bsin⁡xc+dcos⁡x\frac{a+b\sin x}{c+d\cos x} ... (i)
Differentiating (i) with respect to x, we get
ddx\frac{d}{dx} (f (x)) =
(a+bsin⁡x)′(c+dcos⁡x)−(a+bsin⁡x)(c+dcos⁡x)′(c+dcos⁡x)2\frac{(a+b\sin x)'(c+d\cos x)-(a+b\sin x)(c+d\cos x)'}{(c+d\cos x)^2}
=
bcos⁡x(c+dcos⁡x)−(a+bsin⁡x)(−dsin⁡x)(c+dcos⁡x)2\frac{b\cos x(c+d\cos x)-(a+b\sin x)(-d\sin x)}{(c+d\cos x)^2}
=
bccos⁡x+bdcos⁡2x+adsin⁡x+bdsin⁡2x(c+dcos⁡x)2\frac{bc\cos x+bd\cos^2 x+ad\sin x+bd\sin^2 x}{(c+d\cos x)^2}
=
bccos⁡x+bd(cos⁡2x+sin⁡2x)+adsin⁡x(c+dcos⁡x)2\frac{bc\cos x+bd(\cos^2 x+\sin^2 x)+ad\sin x}{(c+d\cos x)^2}
= bccos⁡x+asdsin⁡x+bd(c+dcos⁡x)2\frac{bc\cos x+asd\sin x+bd}{(c+d\cos x)^2}
© examsnet.com
Go to Question: