Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • Differentiate the functions w.r.t. x. x + 3 ^ 2 x + 4 ^3 x + 5 ^ 4

Q5 Differentiate the functions w.r.t. x.   ( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4

 

Answers (1)

best_answer

Given function is
y=( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4
Take log on both sides
\log y=\log [( x+3 )^ 2 . ( x +4 )^ 3 . ( x+5 )^4]\\ \log y = 2\log(x+3)+3\log(x+4)+4\log(x+5)
Now, differentiate w.r.t. x  we get,
\frac{1}{y}.\frac{dy}{dx} = 2.\frac{1}{x+3}+3.\frac{1}{x+4}+4.\frac{1}{x+5}\\ \frac{dy}{dx}=y\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2}{x+3}+\frac{3}{x+4}+\frac{4}{x+5} \right )\\ \frac{dy}{dx} = (x+3)^2.(x+4)^3.(x+5)^4.\left ( \frac{2(x+4)(x+5)+3(x+3)(x+5)+4(x+3)(x+4)}{(x+3)(x+4)(x+5)} \right )\\ \frac{dy}{dx} = (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)
Therefore, the answer is (x + 3) (x + 4)^2 (x + 5)^3 (9x^2 + 70x + 133)
 

Posted by

Gautam harsolia

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

CBSE 12th Result 2026: Board refutes OSM glitch claims, says Class 12 results likely in third week of May
anam-khan

CBSE 12th Result 2026 LIVE: Class 12 scores soon at cbse.gov.in, how to check, digilocker, pass percentage
anam-khan

CBSE 12th Result 2026 Date LIVE: Class 12 marksheet on cbse.gov.in, DigiLocker soon; OSM system, passing marks

Latest Question