Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • <img alt="y=2^{\sin ^{-1} x} \times e^{\cos ^{-1}(x-2)} \text {, then } \frac{d y}{d x} \text { is }" src="https://entrancecorner.oncodecogs.com/gif.latex?y%3D2%5E%7B%5Csin%20%5E%7B-1%7D%20x%7D%20%

y=2^{\sin ^{-1} x} \times e^{\cos ^{-1}(x-2)} \text {, then } \frac{d y}{d x} \text { is }

Option: 1

\frac{\mathrm{y} \log 2}{\sqrt{1-\mathrm{x}^2}}


Option: 2

\frac{y \log (2 / e)}{\sqrt{1-x^2}}


Option: 3

\frac{y-1}{\sqrt{1-x^2}}


Option: 4

none of these


Answers (1)

best_answer

\begin{aligned} & y=2^{\sin ^{-1} x} \cdot e^{\cos ^{-1}(x-2)} \\ & \log y=\sin ^{-1} x \log 2+\cos ^{-1}(x-2) \\ & \frac{1}{y} \cdot \frac{d y}{d x}=\frac{\log 2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-(x-2)^2}}=\frac{\log 2}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2-4+4 x}} \end{aligned}

= answer is none of these.

Hence (D) is correct answer.

 

Posted by

Rishabh

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Marks vs Percentile 2025: How many marks do you need to score 95+ percentile?
anam-khan

JEE Main 2025 session 1 official answer key out; challenge by February 6; how to calculate NTA score?
anam-khan

JEE Main 2025: NIT Raipur's CSE cut-offs drop; closing ranks in popular branches

Latest Question