Derivatives using b^x

Question

I don't understand why, where the lines I noted with red, you would use product rule again?

I know that derivative of b^x is b^x (lnb), but why would you use the product rule on something that you had already take nthe derivative to equal lnb?

What I thought you wre supposed to get was just (8(x^2) + 5) ^(cosx) ln(8(x^2)+5) 16x

What you have is not of the form $b^x$, but of the form $b(x)^{c(x)}$, where $b$ and $c$ are both functions of $x$. So, you need to use the chain rule, in one form or another. — Gerry Myerson, Feb 19 '14 at 06:51

score 1 · Accepted Answer · answered Feb 19 '14 at 16:59

In formulas of the form $f(x)=b(x)^{c(x)}$ you compute the derivative in the most expedient manner via the logarithmic derivative, as it is also used in an explicit manner in your text example,

$\frac{d}{dx}\ln|f(x)|=\frac{f'(x)}{f(x)}$ or $f'(x)=f(x)\cdot \frac{d}{dx}\ln|f(x)|$

Here, $\ln|f(x)|=c(x)\cdot ln|b(x)|$ so that after product rule and chain rule

$$\frac{d}{dx}\ln|f(x)|=c'(x)\cdot \ln|b(x)|+c(x)\cdot \frac{b'(x)}{b(x)}$$

or

$$f'(x)=b(x)^{c(x)}\cdot\left(c'(x)\cdot \ln|b(x)|+c(x)\cdot \frac{b'(x)}{b(x)}\right)$$

Derivatives using b^x

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