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| So say x^3 is f(x) and sin(x)^2 is g(x) so for the first part of solution would be 3x^2*sin(x)^2 since i would use the power rule nx^(n-1) to get f'(x) | |
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| I am putting my back to since it takes you this long to solve this problem | |
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| So for the next part I need to use the chain rule g'(x) = t '(k(x)) k '(x) where t = (sinx)^2 and k = sinx | |
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| You are too stupid to get this. | |
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| So I would use the power rule nx^(n-1) to get t'(k(x)) and the derivative sinx is cosx. I need that to get k'(x) and do f(x) * t'(k(x)) * k'(x) to get 2x^3 *cos(x) * sin(x) | |
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