Help W/ Chain rule using a natural logarithm
- Thread starter Embc
- Start date
Hello, Embc!
I couldn't figure out what you did . . .
I simplified the function first . . .
\(\displaystyle f(x) \;\;=\;\;\ln\bigg[x(2-x^2)^{\frac{1}{2}}\bigg] \;\;=\;\;\ln(x) + \ln(2-x^2)^{\frac{1}{2}} \;\;=\;\;\ln(x) + \tfrac{1}{2}\ln(2-x^2)\)
\(\displaystyle \text{Then: }\;f'(x) \;\;=\;\;\frac{1}{x} + \frac{1}{2}\cdot\frac{-2x}{2-x^2} \;\;=\;\; \frac{1}{x} - \frac{x}{2-x^2} \;\;=\;\;\frac{2-2x^2}{2-x^2}\)
I couldn't figure out what you did . . .
I simplified the function first . . .
\(\displaystyle f(x) \;\;=\;\;\ln\bigg[x(2-x^2)^{\frac{1}{2}}\bigg] \;\;=\;\;\ln(x) + \ln(2-x^2)^{\frac{1}{2}} \;\;=\;\;\ln(x) + \tfrac{1}{2}\ln(2-x^2)\)
\(\displaystyle \text{Then: }\;f'(x) \;\;=\;\;\frac{1}{x} + \frac{1}{2}\cdot\frac{-2x}{2-x^2} \;\;=\;\; \frac{1}{x} - \frac{x}{2-x^2} \;\;=\;\;\frac{2-2x^2}{2-x^2}\)
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- Apr 12, 2005
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What was the question?
h(x) = 2-x^2
j(x) = x^(1/2)
f(x) = ln(x*j(h(x))
\(\displaystyle \frac{df}{dx}\;=\;\frac{1}{x*j(h(x))}\cdot \left[j(h(x))\cdot 1 + x\cdot j'(h(x))\cdot h'(x)\right]\)
You must learn to break them down to tractable pieces. This is one way. The use of the logarithm rules is another excellent siplification.
h(x) = 2-x^2
j(x) = x^(1/2)
f(x) = ln(x*j(h(x))
\(\displaystyle \frac{df}{dx}\;=\;\frac{1}{x*j(h(x))}\cdot \left[j(h(x))\cdot 1 + x\cdot j'(h(x))\cdot h'(x)\right]\)
You must learn to break them down to tractable pieces. This is one way. The use of the logarithm rules is another excellent siplification.