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anonymous
 one year ago
???
anonymous
 one year ago
???

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(f(x) = (x+2)^{7x}\) , right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so, \(y= (x+2)^{7x}\) ln both sides, what do you get?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did chain rule and got 7x(x+2)^(6x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No way!! this is an exponent function with variable on the exponent. You use chain rule if the exponent is a constant. Only one way to take the exponent down is take \(ln\) both sides and use implicit derivative to find dy/dx.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im not sure how to do that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok, I do it for you as sample \(ln y = 7x ln(x+2)\) Now take derivative both sides \(\dfrac{y'}{y}= 7ln(x+2) +\dfrac{7x}{x+2}\) multiple y both sides, and replace \(y = (x+2)^{7x}\) , you get the answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so it would be (7ln(x+2) + 7x/(x+2))(x+2)^7x

phi
 one year ago
Best ResponseYou've already chosen the best response.0oops used implicit differentiation another way is to use this factoid (by definition): \[ x= e^{\ln x } \] use that to say \[ (x+2)= e^{\ln(x+2)} \] and \[ (x+2)^{7x}= \left(e^{\ln(x+2)} \right)^{7x} \\y= e^{7x\ln(x+2)} \] now you can take the deriviative

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much @phi !!!
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