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WebGraphing.com Forum » List all forums » » Forum: Calculus Homework Help » » » Thread: Find F ' (x) |
| Posted by Fabriziolopes at Jan 7, 2008 11:31:14 PM |
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Find F ' (x) F(x) = definite integral from 0 to x^3 of sin t^2 dt F9x) = definite integral from 0 to x^2 sin tetha^2 d tetha ---------------------------------------- Cheese |
| Posted by pskinner at Jan 8, 2008 10:39:30 AM | ||
Re: Find F ' (x)
I'll do the first one; you should be able to get the general idea from understanding that one. The right hand side formula is a composition of functions, f(g(x)), where g(x)=x^3, so you need to use the chain rule to differentiate: F'(x)=f'(g(x))*g'(x) Here, f(x) is the integral from 0 to x of sin(t^2), so f'(x)=sin(x^2) and f'(x^3)=sin((x^3)^2)=sin(x^6). Also, g(x)=x^3 so g'(x)=3x^2. Thus, F'(x)=sin(x^6)*3x^2=3(x^2)sin(x^6) Hope this procedure explains things so you understand what to do with your second one. ---------------------------------------- Principal Skinner |