If y=x+x2+a2n then dydx=

y=x+x2+a2n dydx=nx+x2+a2n-1×1+12x2+a2(2x) =nx+x2+a2n-1·x2+a2+xx2+a2 =n x+x2+a2nx2+a2=nyx2+a2

If y=x+x2+a2n then dydx=

y=x+x2+a2n dydx=nx+x2+a2n-1×1+12x2+a2(2x) =nx+x2+a2n-1·x2+a2+xx2+a2 =n x+x2+a2nx2+a2=nyx2+a2

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

$If y = {(x + \sqrt{x^{2} + a^{2}})}^{n} then \frac{dy}{dx} =$

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

$n y$

$\frac{y}{\sqrt{x^{2} + a^{2}}}$

$\frac{n y}{\sqrt{x^{2} + a^{2}}}$

$y$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$y = {(x + \sqrt{x^{2} + a^{2}})}^{n} \frac{dy}{dx} = n {[x + \sqrt{x^{2} + a^{2}}]}^{n - 1} \times [1 + \frac{1}{2 \sqrt{x^{2} + a^{2}}} (2 x)] = n {[x + \sqrt{x^{2} + a^{2}}]}^{n - 1} \cdot \frac{[\sqrt{x^{2} + a^{2}} + x]}{\sqrt{x^{2} + a^{2}}} = n \frac{{[x + \sqrt{x^{2} + a^{2}}]}^{n}}{\sqrt{x^{2} + a^{2}}} = \frac{ny}{\sqrt{x^{2} + a^{2}}}$