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The function f is defined by f(x)=the sqr root 25-x squared for -5 <(or equal to) x<( or equal to) 5. Find f'(x) Write an equation for the line tangent to the graph of f at x=-3. and 2 other questions
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\[f(x)=\sqrt{25-x^2}\]?
That is correct
\[\large \frac{d}{dx}[\sqrt{g(x)}]=\frac{g'(x)}{2\sqrt{g(x)}}\] by the chain rule in your case \(g(x)=25-x^2\)
so i take the derivative of it and I do that by using the chain rule is what you're saying?
yes
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okay does this seem right? |dw:1463539466992:dw|
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