If f(x)=log(x2+x+1)+cosh(2x−3),then f'(0)=

f(x)=log⁡x2+x+1+cosh⁡(2x−3)f(x)=log⁡x2+x+1+cosh⁡(2x−3)1/21/2f′(x)=12log⁡x2+x+1+cosh⁡(2x−3)1/2ddxlog⁡x2+x+1+ddxcosh⁡(2x−3)1/2=12log⁡x2+x+1+cosh⁡(2x−3)1/21/22x+1x2+x+1+12cosh⁡(2x−3)−1/2×sinh⁡(2x−3)×2f′(0)=12[log⁡(0+1+0)+cosh⁡(−3)1/2−1/211+12cosh⁡(−3)−1/2×sinh⁡(−3)×2=120+(cosh⁡3)1/2−1/2×[1+cosh⁡3)−1/2×(−sinh⁡3)]=12(cosh⁡3)−1/41−sinh⁡3(cosh⁡3)1/2=cosh⁡(3)−sinh⁡(3)2(cosh⁡(3))3/4

If f(x)=log(x2+x+1)+cosh(2x−3),then f'(0)=

f(x)=log⁡x2+x+1+cosh⁡(2x−3)f(x)=log⁡x2+x+1+cosh⁡(2x−3)1/21/2f′(x)=12log⁡x2+x+1+cosh⁡(2x−3)1/2ddxlog⁡x2+x+1+ddxcosh⁡(2x−3)1/2=12log⁡x2+x+1+cosh⁡(2x−3)1/21/22x+1x2+x+1+12cosh⁡(2x−3)−1/2×sinh⁡(2x−3)×2f′(0)=12[log⁡(0+1+0)+cosh⁡(−3)1/2−1/211+12cosh⁡(−3)−1/2×sinh⁡(−3)×2=120+(cosh⁡3)1/2−1/2×[1+cosh⁡3)−1/2×(−sinh⁡3)]=12(cosh⁡3)−1/41−sinh⁡3(cosh⁡3)1/2=cosh⁡(3)−sinh⁡(3)2(cosh⁡(3))3/4

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 $f (x) = \sqrt{\log (x^{2} + x + 1) + \sqrt{\cosh (2 x - 3)}},$ then $f^{'} (0) =$

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

$\frac{\log 3 \sqrt{\cosh (3)} - \sinh (3)}{2 {(\cosh (3))}^{\frac{3}{4}}}$

$\frac{1}{2 \sqrt{\sqrt{\cosh (3)}}} (\log 3 - \frac{\sinh (3)}{\sqrt{\cosh (3)}})$

$\frac{1}{2 \sqrt{\sqrt{\cosh (3)}}} (1 + \frac{\sinh (3)}{\sqrt{\cosh (3)}})$

$\frac{\sqrt{\cosh (3)} - \sinh (3)}{2 {(\cosh (3))}^{\frac{3}{4}}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} f (x) = \sqrt{\log (x^{2} + x + 1) + \sqrt{\cosh (2 x - 3)}} \\ f (x) = [\log (x^{2} + x + 1) + {(\cosh (2 x - 3)^{1 / 2}]}^{1 / 2} \\ f^{'} (x) = \frac{1}{2} [\log (x^{2} + x + 1) + (\cosh (2 x - 3)^{1 / 2} \\ [\frac{d}{d x} \log (x^{2} + x + 1) + \frac{d}{d x} (\cosh (2 x - 3)^{1 / 2}] \\ = \frac{1}{2} (\log (x^{2} + x + 1) + {(\cosh (2 x - 3)^{1 / 2})}^{1 / 2} \\ [\frac{2 x + 1}{x^{2} + x + 1} + \frac{1}{2} (\cosh (2 x - 3)^{- 1 / 2} \times \sinh (2 x - 3) \times 2] \\ f^{'} (0) = \frac{1}{2} [\log (0 + 1 + 0) \\ + {(\cosh (- 3)^{1 / 2}]}^{- 1 / 2} \\ [\frac{1}{1} + \frac{1}{2} (\cosh (- 3)^{- 1 / 2} \times \sinh (- 3) \times 2] \\ = \frac{1}{2} {[0 + (\cosh 3)^{1 / 2}]}^{- 1 / 2} \times [1 + \cosh 3)^{- 1 / 2} \\ \times (- \sinh 3)] \end{array}$
$= \frac{1}{2} (\cosh 3)^{- 1 / 4} (1 - \frac{\sinh 3}{(\cosh 3)^{1 / 2}})$
$= \frac{\sqrt{\cosh (3)} - \sinh (3)}{2 (\cosh (3))^{3 / 4}}$