If f(x)= 7x+1sinxexlogxand f'(x)=f(x)g(x), then g'(x)=

f(x)=(7x+1)sinxex logx log f(x)=log (7x+1)+log sinx-log ex-log(logx) f1(x)f(x)=77x+1+cot x-1-1x log xf1(x)=f(x)77x+1+cot x-1-1x log x g(x)=77x+1+cot x-1-1x logx g'(x)=-49(7x+1)2-cosec2x+1(x logx)2(1+logx)

If f(x)= 7x+1sinxexlogxand f'(x)=f(x)g(x), then g'(x)=

f(x)=(7x+1)sinxex logx log f(x)=log (7x+1)+log sinx-log ex-log(logx) f1(x)f(x)=77x+1+cot x-1-1x log xf1(x)=f(x)77x+1+cot x-1-1x log x g(x)=77x+1+cot x-1-1x logx g'(x)=-49(7x+1)2-cosec2x+1(x logx)2(1+logx)

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

If f(x)= $\frac{(7 x + 1) \sin x}{e^{x} \log x}$ and f'(x)=f(x)g(x), then g'(x)=

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{{(x \log x)}^{2}} + \frac{x}{\log x} - \csc^{2} x - \frac{49}{{(7 x + 1)}^{2}}$

$\frac{1}{x^{2} \log x} + \frac{1}{{(x \log x)}^{2}} - \csc^{2} x + \frac{49}{{(7 x + 1)}^{2}}$

$\frac{1}{x^{2} \log x} + \frac{1}{{(x \log x)}^{2}} - \csc^{2} x - \frac{49}{{(7 x + 1)}^{2}}$

$\frac{1}{x^{2} \log x} + \frac{1}{\log x} - \csc^{2} x - \frac{49}{{(7 x + 1)}^{2}}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$f (x) = \frac{(7 x + 1) sinx}{e^{x} logx} \log f (x) = \log (7 x + 1) + \log sinx - \log e^{x} - \log (logx) \frac{f^{1} (x)}{f (x)} = \frac{7}{7 x + 1} + \cot x - 1 - \frac{1}{x \log x}$

$f^{1} (x) = f (x) [\frac{7}{7 x + 1} + \cot x - 1 - \frac{1}{x \log x}] g (x) = \frac{7}{7 x + 1} + \cot x - 1 - \frac{1}{x logx} g' (x) = \frac{- 49}{{(7 x + 1)}^{2}} - {cosec}^{2} x + \frac{1}{{(x logx)}^{2}} (1 + logx)$