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I am lost on deriving this one function, because im kind of confused with the $9e^x$ and the fraction part. Maybe if someone can guide me through the steps that would be awesome.

$$y=9e^x+\frac{2}{\sqrt[3]{x}}$$

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  • $\begingroup$ Is this homework? $\endgroup$ – VF1 Oct 8 '12 at 23:58
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Some rules that may help you:

For a function $f(x)$ and its derivative with respect to x $f'(x)$, $$ \frac{d}{dx} c f(x) = c f'(x) $$

You may want to look up what the derivative of the exponent function is. That's an important one to know. Also, for the fraction part, consider that: $$ \frac{2}{\sqrt[3]{x}} = \frac{2}{x^{1/3}} = 2 x^{-1/3} $$ For this one you may want to look up the power rule.

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This is a rather cook-book answer, but. . .

As said above, use that $\frac{d}{dx}cf(x)=c\frac{d}{dx}f(x)$ and $\frac{2}{\sqrt[3]{x}}=2x^{-\frac{1}{3}}.$

Follow that up with the fact that $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}x^m=mx^{m-1}.$

The first follows from many things (particularly the definition of $e^x$). The second follows from the definition of the derivative.

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Just for instruction's sake, I'm going to solve this with logarithms. Bear in mind:

$$\log(a \cdot b) = \log a + \log b$$

and

$$\log(a^x) = x \log a$$

and

$$\log e = 1$$

Let $y_1 = 9e^x$ and $y_2 = 2x^{-1/3}$. Now,

$$\log y_1 = \log 9 + x$$

Differentiating w.r.t $x$,

$$\frac{1}{y_1} \frac{dy_1}{dx} = 1$$ Thus, $ \frac{dy_1}{dx} = y_1 = 9e^x$ (which is not surprising since that's a property of the exponential function).

Now, $\log y_2 = \log 2 -\frac{1}{3} \log x$. Hence, differentiating w.r.t $x$,

$$\frac{1}{y_2} \frac{dy_2}{dx} = -\frac{1}{3} \frac{1}{x}$$

Thus, $$\frac{dy_2}{dx} = -\frac{2}{3} x^{-4/3}$$

And therefore, $$\frac{dy}{dx} = 9e^x -\frac{2}{3} x^{-4/3}$$

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