`y=x^4/cosx`

Using the quotient rule for evaluating the derivative of the function,

`y'=(cosxd/dx(x^4)-x^4d/dxcosx)/(cosx)^2`

`y'=(cos(x)*(4x^3)-x^4(-sin(x)))/(cos^2(x))`

`y'=(4x^3cos(x)+x^4sin(x))/(cos^2(x))`

`y'=(x^3(4cos(x)+xsin(x)))/(cos^2(x))`

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`y=x^4/cosx`

Using the quotient rule for evaluating the derivative of the function,

`y'=(cosxd/dx(x^4)-x^4d/dxcosx)/(cosx)^2`

`y'=(cos(x)*(4x^3)-x^4(-sin(x)))/(cos^2(x))`

`y'=(4x^3cos(x)+x^4sin(x))/(cos^2(x))`

`y'=(x^3(4cos(x)+xsin(x)))/(cos^2(x))`

You need to evaluate the derivative of the given function and since the function is a quotient of two functions, then you must use the quotient rule, such that:

`f'(x) = ((x^4)'(cos x) - (x^4)(cos x)')/((cos x)^2)`

`f'(x) = (4x^3*cos x + x^4*sin x)/((cos x)^2)`

Factoring out `x^3` yields:

`f'(x) = x^3(4x*cos x + x*sin x)/((cos x)^2)`

**Hence, evaluating the derivative of the function, using the product rule, yields `f'(x) = x^3(4x*cos x + x*sin x)/((cos x)^2).` **