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Calculus1 16 Online
OpenStudy (anonymous):

derivative of f(x)=sin^2(x^3+3) PLEASE HELP

OpenStudy (hlares):

Alright, so, taking the derivative of f(x)=(sin(x^3+3))^2. When it comes to working with derivatives like that, I like to think of them like nesting dolls where I either begin with the outermost part of the function (the exponent 2 in this case) or the innermost (x^3+3 in this case) and then work inwards or outwards from there. Going from the innermost and out: What you will need to do is take the derivative of the innermost function (x^3+3), then the next function (sin(x^3+3), and then finally take the derivative of the exponent. After that, find the product of the derivatives/simplify what you have. The individual steps: Derivative of f(x)=x^3+3 is f'(x)=3x^2 +0=3x^2 Derivative of f(x)=sin(x^3+3)=cos(x^3+3) Derivative of f(x)=(sin(x^3+3)^2=2(sin(x^3+3) And all together: f(x)=(sin(x^3+3))^2 f'(x)=3x^2*cos(x^3+3)*2sin(x^3+3) f'(x)=6x^2*sin(x^3+3)*cos(x^3+3)

OpenStudy (hlares):

Another tip for working on derivatives like this: when you are taking the derivative of anything but the innermost function, do not focus on what is inside the parentheses. Instead, just pretend it's a random variable. For example: with sin(x^3+3) I just think of it as sin(x) and take the derivative of that and with (sin(x^3+3))^2 I just think of it like x^2 and take the derivative of that. It helps in focusing on the step and in not getting distracted by or including variables you do not need to worry about for that step.

OpenStudy (anonymous):

This is amazing thank you @Hlares

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