Find the derivative of function y=(sin(cos(2pix^3)))*(1/x^2)?

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You need to use the product, quotient and chain rules of derivatives.

`dy/dx = (sin(cos(2pix^3)))'*(1/x^2) + (sin(cos(2pix^3)))*(1/x^2)'`

You need to use chain rule to differentiate`sin(cos(2pix^3)).`

`sin(cos(2pix^3))' = cos(cos(2pix^3))(cos(2pi*x^3))'*(2pi*x^3)'`

`sin(cos(2pix^3))' = -(cos(cos(2pix^3)))*(sin(2pi*x^3))*(6pi*x^2)`

You need to use product rule to differentiate `1/x^2` .

`(1/x^2)' = ((1')*(x^2) - 1*(x^2)')/x^4`

`(1/x^2)' = (0-2x)/x^4 =gt (1/x^2)' = -2x/x^4 =gt (1/x^2)' = -2/x^3`

`dy/dx = -((cos(cos(2pix^3)))*(sin(2pi*x^3))*(6pi*x^2))/x^2 - 2* (sin(cos(2pix^3)))/x^3`

**The derivative of the function `y=(sin(cos(2pix^3)))*(1/x^2)` is `dy/dx = 6pi*(cos(cos(2pix^3)))*(sin(2pi*x^3))- 2 (sin(cos(2pix^3)))/x^3` **

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