# Differentiation: Finding the Derivative

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In solving the problem, it is determined that the derivative of h(x) = sin^2x + cosx.

0 is less than x which is less than 2pi

was: h'(x) = cosxsinx +sinxcosx - sinx which then simplified to 2sinxcosx-sinx which then simplified to sinx(2cosx-1)

When do the derivative, I get cos^2x -sinx

Would you explain what I am doing wrong in finding the derivative if you can figure that out, and would you please provide a step by step analysis of how the you got the derivative above.

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#### Solution Preview

Your mistake is, when you do derivative for sin^2x, you get ((sinx)')^2=(cosx)^2=cos^2x. You simply exchange the order of square and sin.

f(x)=sin^2x+cosx, ...

#### Solution Summary

The solution assists with showing the steps to finding the derivative.

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