What is the derivative of? : y=3tan^(-1)(x+sqrt(1+x^2) ) y=3tan−1(x+√1+x2)
1 Answer
dy/dx = ( 3(1+x/(sqrt(1+x^2))))/(1+(x+sqrt(1+x^2))^2) dydx=3(1+x√1+x2)1+(x+√1+x2)2
Explanation:
We seek (I assume):
dy/dx dydx wherey=3tan^(-1)(x+sqrt(1+x^2) ) y=3tan−1(x+√1+x2)
We will need the standard result:
d/dx tan^(-1)x = 1/(1+x^2) ddxtan−1x=11+x2
And the power rule, in conjunction with the chain rule, for:
d/dx (x+ sqrt(1+x^2) ) = 1+d/dx (1+x^2)^(1/2)ddx(x+√1+x2)=1+ddx(1+x2)12
" " = 1+1/2(1+x^2)^(-1/2) d/dx (1+x^2) =1+12(1+x2)−12ddx(1+x2)
" " = 1+1/(2sqrt(1+x^2)) (2x) =1+12√1+x2(2x)
" " = 1+x/(sqrt(1+x^2)) =1+x√1+x2
Then we can apply the product rule to the original function to get:
dy/dx = 3 {1/(1+(x+sqrt(1+x^2))^2) } d/dx (x+sqrt(1+x^2) ) dydx=3⎧⎪ ⎪⎨⎪ ⎪⎩11+(x+√1+x2)2⎫⎪ ⎪⎬⎪ ⎪⎭ddx(x+√1+x2)
\ \ \ \ \ = 3 {1/(1+(x+sqrt(1+x^2))^2) } {1+x/(sqrt(1+x^2))}
\ \ \ \ \ = ( 3(1+x/(sqrt(1+x^2))))/(1+(x+sqrt(1+x^2))^2)
