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If #f(x) =-e^(-3x-7) # and #g(x) = (lnx)^2 #, what is #f'(g(x)) #?

Answer 1

Isaiah Allen

See derivation below

First of all f(g(x) would be as follows, which can be differentiated using chain rule.

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Answer 2

Aurora Alvarado

To find ( f'(g(x)) ), we first need to find ( g'(x) ) and then substitute it into the derivative of ( f ).

First, let's find ( g'(x) ): [ g(x) = (\ln x)^2 ] [ \frac{d}{dx}g(x) = 2(\ln x)\frac{1}{x} = \frac{2\ln x}{x} ]

Now, let's find ( f'(g(x)) ): [ f(g(x)) = -e^{-3g(x)-7} ] [ \frac{d}{dx}f(g(x)) = -e^{-3g(x)-7}(-3g'(x)) ] [ = 3e^{-3(\ln x)^2 - 7} \frac{2\ln x}{x} ] [ = 3\frac{2\ln x}{xe^{3(\ln x)^2 + 7}} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.