Limits Continuity And Differentiability Question 61
Question: If $ u=f(x^{3}),v=g(x^{2}),f’(x)=\cos x $ and $ g’(x)=\sin x, $ then $ \frac{du}{dv}= $
Options:
A) $ \frac{1}{2}x\cos x^{3}\cos ecx^{2} $
B) $ \frac{3}{2}x\cos x^{3}\cos ecx^{2} $
C) $ \frac{1}{2}x\sec x^{3}\sin x^{2} $
D) $ \frac{3}{2}x\sec x^{3}\cos ecx^{2} $
Show Answer
Answer:
Correct Answer: B
Solution:
Here, $ u=f(x^{3}) $
$ \Rightarrow \frac{du}{dx}=f’(x^{3}).\frac{d}{dx}(x^{3}) $
$ =(\cos (x^{3})).3x^{2}=3x^{2}.\cos x^{3} $ and $ v=g(x^{2}) $
$ \Rightarrow \frac{dv}{dx}=g’(x^{2}).\frac{d}{dx}(x^{2})=(\sin x^{2}).(2x) $
$ =2x\sin x^{2} $
$ \therefore \frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}\frac{3x^{2}.\cos x^{2}}{2x.\sin x^{2}} $
$ \Rightarrow \frac{du}{dv}=\frac{3}{2}x.\cos x^{3}.\cos ecx^{2} $
BETA
