I am stuck in these problems.
I think for the first problem the answer is $\dfrac{2}{x^7}$, whereas for the second problem the answer is $x$.
These two are pretty simple, so I just want to make sure I got the hang of it. Thanks =]
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It seems that you are not getting the hang of solving such problems. It really helps learning AND applying the derivative identities correctly. Here is how you approach the problems you are working on.
$$\Large\textbf{First Problem}$$
We need to determine the derivative of $\log_2(x^8)$. Observe that by the properties of logarithm, it is equivalent to $$\dfrac{\log(x^8)}{\log(2)} = \dfrac{8\log(x)}{\log(2)}$$
Since
$$\dfrac{d}{dx}[\log(x)] = \dfrac{1}{x}$$
we have
$$\dfrac{d}{dx}\left[\log_2(x^8)\right] = \dfrac{d}{dx}\left[\dfrac{8\log(x)}{\log(2)} \right] = \dfrac{8}{\log(2)} \dfrac{d}{dx}[\log(x)] = \dfrac{8}{\log(2)}\cdot \dfrac{1}{x}$$
$$\Large\textbf{Second Problem}$$
You need to use Product Rule, which states that if $f$ and $g$ are differentiable functions, then
$$\dfrac{d}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$$
First, let $f(x) = e^x$ and $g(x) = \ln(x)$. Then, $f'(x) = e^x$ and $g'(x) = \frac{1}{x}$. Thus,
$$\dfrac{d}{dx}[e^x\ln(x)] = e^x\cdot \ln(x) + e^x \cdot \dfrac{1}{x} = e^x\ln(x) + \dfrac{e^x}{x}$$
For the first one, notice that from the properties of logarithms, $\log_2{x^8} = \frac{\ln{x^8}}{\ln{2}} = \frac{8}{\ln{2}}\cdot\ln{x}$. Differentiating this should be easy.
For the second one, apply the product rule : Let $v = e^x$ and $u = \ln{x}$. Then $$\frac{d}{dx}(uv) = u\cdot\frac{dv}{dx} + v\cdot\frac{du}{dx}$$ which should be rather simple as well!