f(x)= 10x^(lnx)
How do you find the value of the derivative when x =8?
You have: $\displaystyle f(x) = 10x^{ln(x)}$
Apply the chain rule:
$\displaystyle
10x^{ln (x)} \frac{d}{dx} (ln^2x)
$
u=log x and n=2
$\displaystyle
= 20x^{ln(x)} ln(x) \frac{d}{dx} (ln(x))
$ and since the derivative of ln(x) = 1/x
you have: $\displaystyle f'(x) = 20 x^{ln(x)-1}ln(x)$
Your answer is correct but I cannot follow the logic.
Using logarithmic differentiation:
$\displaystyle y = 10 x^{\ln x} \Rightarrow \ln y = (\ln x)^2 + \ln 10$.
Therefore $\displaystyle \frac{1}{y} \frac{dy}{dx} = \frac{2 \ln x}{x} \Rightarrow \frac{dy}{dx} = \frac{2 y \ln x}{x} = \frac{2 \left(10 x^{\ln x}\right) \ln x}{x} = 20 x^{\ln (x) - 1} \ln x$.
Also, just in case a picture helps convey the logic that Roam (possibly) and - Wolfram|Alpha (definitely) are using...
... where
... is the chain rule for two inner functions, i.e...
$\displaystyle \frac{d}{dx}\ f(u(x), v(x)) = \frac{\partial f}{\partial u} \frac{du}{dx} + \frac{\partial f}{\partial v} \frac{dv}{dx}$
As with...
... the ordinary chain rule, straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the (corresponding) dashed balloon expression which is (one of) the inner function(s) of the composite expression.
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