0

Guest

Courses New

y = a^cos-1x/1+a^cos-1x. z=a^cos-1x,then dy/dx ?A.1/1+a^cos-1x. B.-1/a^cos-1x C.(1/1+ a^cos-1)^2

y = a^cos-1x/1+a^cos-1x. z=a^cos-1x,then dy/dx ?A.1/1+a^cos-1x. B.-1/a^cos-1x C.(1/1+ a^cos-1)^2

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
Dear Student,
the question must have asked dy/dz as z value is given.
y=a^cos-1x/(1+a^(cos^-1 x)) and z=a^(cos^-1 x)
 
dy/dx=( -a^(cos^-1 x)*log a*1*(1+a^(cos^-1 x))/(sqrt(1-x^2)) +
a^(cos^-1 x)*log a*a^(cos^-1 x)/(sqrt(1-x^2)))/(1+a^(cos^-1 x))^2
=((a^(cos^-1 x)*log a)*(-1-a^(cos^-1 x)+a^(cos^-1 x)))/((sqrt(1-x^2)*(1+a^(cos^-1 x)^2)
 
now, dz/dx =-a^(cos^-1 x)*log a /(sqrt(1-x^2))
so dy/dz= (dy/dx)/(dz/dx)= 1/(1+a^(cos^-1 x))^2
 
So C is the correct answer.
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)

Think You Can Provide A Better Answer ?

Other Related Questions on Integral Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More