1. find f'(x) of
I got using the logarithmic rule and the product rule right?
2. If , find .
What do I do first? Take the ln so I get the ^x in front of the ln? Thanks
1. The right direction - you've basically done...
... where (key in spoiler) ...
Spoiler:
... but you've not finished solving the bottom row for f'(x)...
7 ln x you've multiplied by f(x) but not the other term (the one cancelling down to 7).
Yes - so you could draw a similar picture?
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Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
whenever you are told to differentiate something like in Q1 and Q2, take the natural logarithm of both sides.
$\displaystyle y = x^{7x}$
take natural logarithm
$\displaystyle ln y = 7x . lnx$
differentiate implicitly. on RHS, use product rule
$\displaystyle \frac{y'}{y} = 7 lnx + 7x . \frac{1}{x} $
$\displaystyle y' = x^{7x}.(7lnx + 7) $
remember that when you multiply by y over to the other side, 7lnx and 7x both get multiplied
Hello, softballchick!
$\displaystyle \text{2. If }f(x) \:=\:8(\sin x)^x,\:\text{ find }f'(2)$
$\displaystyle \text{Let }\,y \;=\;8(\sin x)^x$
$\displaystyle \text{Take logs: }\:\ln y \:=\:\ln\left[8(\sin x)^x\right] \;=\;\ln 8 + \ln(\sin x)^x \;=\;\ln8 + x\ln(\sin x) $
$\displaystyle \text{Differentiate implicitly: }\:\dfrac{1}{y}\cdot y' \;=\;x\cdot\dfrac{\cos x}{\sin x} + \ln(\sin x) $
. . . $\displaystyle \dfrac{y'}{y} \;=\;x\cdot\cot x + \ln(\sin x)$
. . . $\displaystyle y' \;=\;y\left[x\cdot\cot x + \ln(\sin x)\right] $
.$\displaystyle f'(x) \;=\;8(\sin x)^x\left[x\cdot\cot x + \ln(\sin x)\right] $
$\displaystyle \text{Therefore: }\:f'(2) \;=\;8(\sin 2)^2\left[2\cot 2 + \ln(\sin 2)\right] \;\approx\;-6.683353787 $