What is derivative 5x^2*cos(arcsin(1-x^2))?

Asked on by clamek

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use product rule and then you need to use chain rule such that:

`(5x^2*cos(arcsin(1-x^2)))' = (5x^2)'*cos(arcsin(1-x^2)) + 5x^2*(cos(arcsin(1-x^2)))'`

`(5x^2*cos(arcsin(1-x^2)))' = 10x*cos(arcsin(1-x^2)) + 5x^2*(-sin(arcsin(1-x^2))*(arcsin(1-x^2))'`

You need to remember that `sin(arcsin x) = x`

`(5x^2*cos(arcsin(1-x^2)))' = 10x*cos(arcsin(1-x^2)) - 5x^2*(1-x^2)*(((1-x^2)')/sqrt(1 - (1-x^2)^2))`

`(5x^2*cos(arcsin(1-x^2)))' = 10x*cos(arcsin(1-x^2))+ (10x^3*(1-x^2))/sqrt((1-1+x^2)(1+1-x^2))`

`(5x^2*cos(arcsin(1-x^2)))' = 10x*cos(arcsin(1-x^2)) + (10x^3*(1-x^2))/sqrt(2x^2-x^4)`

Hence, evaluating derivative of function yields `(5x^2*cos(arcsin(1-x^2)))' = 10x*cos(arcsin(1-x^2)) + (10x^3*(1-x^2))/sqrt(2x^2-x^4).`

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