what is the derivative of this function

Question

I need help in this following question. I have tried many attempts but really confused on how to solve it. Therefore, i would appreciate any help from you guys. Thanks allot

$$f(x) = \dfrac{\ln x}{e^{x^2 + 2}}$$

Answer 1

$$f(x) = \dfrac{\ln x}{e^{x^2 + 2}}$$

We use the quotient rule: $$f(x) = \dfrac {g(x)}{h(x)} \implies f'(x) = \dfrac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}$$

Here, $$g(x) = \ln x \implies g'(x) = \dfrac 1x$$ $$h(x) = e^{x^2 + 2} \implies h'(x) = 2x\cdot e^{x^2 + 2}$$

Note we need the chain rule for finding $h'(x)$.

Can you take it from here? You'll be able to simplify once you "plug into" the formula given by the quotient rule.

$$\dfrac{\frac 1x \cdot e^{x^2 + 2} - \ln x(2xe^{x^2+ 2})}{(e^{x^2 + 2})^2} = \dfrac{e^{x^2 + 2}\Big(\frac 1x - 2x\ln x\Big)}{e^{2(x^2 + 2)}} = \dfrac{1- 2x^2 \ln x}{xe^{x^2 + 2}}$$

Answer 2

Hint: Rewrite the function as:

$$f(x) = \ln x e^{-x^2-2}$$

Now, use the product rule to find the result.

$$\dfrac{d}{dx} ( uv) = v \dfrac{du}{dx} + u \dfrac{dv}{dx}$$

Spoiler

f'(x) = $-2e^{-x^2-2}~x~\ln x + \dfrac{e^{-x^2-2}}{x}$

Of course, you could have also just used the quotient and should give that a go!

Answer 3

another occasionally useful device is to rewrite as: $$(e^{x^2 + 2})f(x) = ln \;x$$ and use the product rule on the LHS to get: $$(2xe^{x^2+2})f(x) + (e^{x^2 + 2})f'(x) = \frac 1x $$ so $$ f'(x) = (\frac1x -2x \;ln \;x)(e^{x^2 + 2})^{-1} $$