`y = arccos((b + acos(x))/(a + bcos(x))), 0<= x<= pi, a>b>0` Find the derivative of the function. Simplify where possible.

Textbook Question

Chapter 3, 3.5 - Problem 59 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`y=cos^-1((b+acos(x))/(a+bcos(x)))` 

`y'=((-1)/sqrt(1-((b+acos(x))/(a+bcos(x)))^2))* d/(dx) ((b+acos(x))/(a+bcos(x)))`

`y'=-(a+bcos(x))/sqrt((a+bcos(x))^2-(b+acos(x))^2) *d/(dx) (b+acos(x))(a+bcos(x))^-1`

`y'=(-(a+bcos(x)))/sqrt((a^2+b^2cos^2(x) +2abcos(x)) -(b^2+a^2cos^2(x) +2abcos(x))) * ((b+acos(x))(-1)(a+bcos(x))^(-2)(-bsin(x)) + (a+bcos(x))^(-1)(-asin(x)))`

`y'=(-(a+bcos(x)))/sqrt(a^2+b^2cos^2(x)-b^2-a^2cos^2(x)) *(((bsin(x))(b+acos(x)))/(a+bcos(x))^2 + (-asin(x))/((a+bcos(x))))`

`y'=(-(a+bcos(x)))/sqrt((a^2-b^2-(a^2-b^2)cos^2(x))) *((bsin(x)(b+acos(x)) -(asin(x))(a+bcos(x)))/(a+bcos(x))^2)`

`y'=(-(a+bcos(x)))/sqrt((a^2-b^2)(1-cos^2(x))) *((b^2sin(x) + a*bsin(x)cos(x) -a^2sin(x)- a*bsin(x)cos(x))/(a+bcos(x))^2)`

`y'=(-(b^2-a^2)sin(x))/((a+bcos(x))(sqrt((a^2-b^2)sin^2(x))))`

`y'=((a^2-b^2)sin(x))/((sin(x)(a+bcos(x))sqrt(a^2-b^2)))`

`y'=(a^2-b^2)/(sqrt(a^2-b^2) (a+bcos(x)))`

`y'=sqrt(a^2-b^2)/(a+bcos(x))`

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