# Help on log derivation

• Nov 19th 2010, 02:54 PM
softballchick
Help on log derivation
1. find f'(x) of

http://math.webwork.rochester.edu:80...e235aa1d41.png

I got http://math.webwork.rochester.edu:80...3268553f61.png using the logarithmic rule and the product rule right?

2. If http://math.webwork.rochester.edu:80...9dab67a7d1.png, find http://math.webwork.rochester.edu:80...eff830eef1.png.

What do I do first? Take the ln so I get the ^x in front of the ln? Thanks

• Nov 20th 2010, 01:33 AM
tom@ballooncalculus
1. The right direction - you've basically done...

http://www.ballooncalculus.org/draw/diffLog/one.png

... where (key in spoiler) ...

Spoiler:
http://www.ballooncalculus.org/asy/chain.png

... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to the main variable (in this case x), and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule). And...

http://www.ballooncalculus.org/asy/prod.png

... is the product rule, straight continuous lines still differentiating downwards with respect to x.

... 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).

Quote:

Originally Posted by softballchick
2. If http://math.webwork.rochester.edu:80...9dab67a7d1.png, find http://math.webwork.rochester.edu:80...eff830eef1.png.

What do I do first? Take the ln so I get the ^x in front of the ln? Thanks

Yes - so you could draw a similar picture?
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• Nov 20th 2010, 03:26 AM
HallsofIvy
Quote:

Originally Posted by softballchick
1. find f'(x) of

No, that's NOT right. How did you get it?

Quote:

2. If http://math.webwork.rochester.edu:80...9dab67a7d1.png, find http://math.webwork.rochester.edu:80...eff830eef1.png.

What do I do first? Take the ln so I get the ^x in front of the ln? Thanks
Quote:

ln(f(x))= ln(8)+ xln(sin(x))
• Nov 20th 2010, 03:57 AM
differentiate
whenever you are told to differentiate something like in Q1 and Q2, take the natural logarithm of both sides.

$y = x^{7x}$
take natural logarithm
$ln y = 7x . lnx$
differentiate implicitly. on RHS, use product rule
$\frac{y'}{y} = 7 lnx + 7x . \frac{1}{x}$
$y' = x^{7x}.(7lnx + 7)$

remember that when you multiply by y over to the other side, 7lnx and 7x both get multiplied
• Nov 20th 2010, 09:55 AM
Soroban
Hello, softballchick!

Quote:

$\text{2. If }f(x) \:=\:8(\sin x)^x,\:\text{ find }f'(2)$

$\text{Let }\,y \;=\;8(\sin x)^x$

$\text{Take logs: }\:\ln y \:=\:\ln\left[8(\sin x)^x\right] \;=\;\ln 8 + \ln(\sin x)^x \;=\;\ln8 + x\ln(\sin x)$

$\text{Differentiate implicitly: }\:\dfrac{1}{y}\cdot y' \;=\;x\cdot\dfrac{\cos x}{\sin x} + \ln(\sin x)$

. . . $\dfrac{y'}{y} \;=\;x\cdot\cot x + \ln(\sin x)$

. . . $y' \;=\;y\left[x\cdot\cot x + \ln(\sin x)\right]$

. $f'(x) \;=\;8(\sin x)^x\left[x\cdot\cot x + \ln(\sin x)\right]$

$\text{Therefore: }\:f'(2) \;=\;8(\sin 2)^2\left[2\cot 2 + \ln(\sin 2)\right] \;\approx\;-6.683353787$