Differentiate problem
- Thread starter hank
- Start date
Product rule:
\(\displaystyle \L\\4x(4x^{2}-1)^{\frac{-1}{2}}\)
\(\displaystyle \L\\4x(\frac{-1}{2})(4x^{2}-1)^{\frac{-3}{2}}(8x)+(4x^{2}-1)^{\frac{-1}{2}}(4)\)
\(\displaystyle \L\\\frac{-16x^{2}}{(4x^{2}-1)^{\frac{3}{2}}}+\frac{4}{(4x^{2}-1)^{\frac{1}{2}}}\)
Multiply right side by (4x2−1)
\(\displaystyle \L\\\frac{-16x^{2}}{(4x^{2}-1)^{\frac{3}{2}}}+\frac{4}{(4x^{2}-1)^{\frac{1}{2}}}\cdot\frac{(4x^{2}-1)}{(4x^{2}-1)}\)
\(\displaystyle \H\\\fbox{\frac{-4}{(4x^{2}-1)^{\frac{3}{2}}}}\)
\(\displaystyle \L\\4x(4x^{2}-1)^{\frac{-1}{2}}\)
\(\displaystyle \L\\4x(\frac{-1}{2})(4x^{2}-1)^{\frac{-3}{2}}(8x)+(4x^{2}-1)^{\frac{-1}{2}}(4)\)
\(\displaystyle \L\\\frac{-16x^{2}}{(4x^{2}-1)^{\frac{3}{2}}}+\frac{4}{(4x^{2}-1)^{\frac{1}{2}}}\)
Multiply right side by (4x2−1)
\(\displaystyle \L\\\frac{-16x^{2}}{(4x^{2}-1)^{\frac{3}{2}}}+\frac{4}{(4x^{2}-1)^{\frac{1}{2}}}\cdot\frac{(4x^{2}-1)}{(4x^{2}-1)}\)
\(\displaystyle \H\\\fbox{\frac{-4}{(4x^{2}-1)^{\frac{3}{2}}}}\)
Hello, Hank!
Galactus is absolutely correct . . .
If there is a constant coefficient, I prefer to "leave it out front".
We have: \(\displaystyle \L\:f(x)\;=\;4\,\cdot\,\frac{x}{\left(4x^2-1\right)^{1/2}}\)
Then: \(\displaystyle \L f'(x) \;= \;4\,\left[\frac{(4x^2-1)^{\frac{1}{2}}\cdot1 \,- \,x\cdot\frac{1}{2}\cdot\left(4x^2-1\right)^{-\frac{1}{2}}\cdot8x}{4x^2-1} \right]\)
. . . . . . . . . . .\(\displaystyle \L=\;4\,\left[\frac{(4x^2-1)^{\frac{1}{2}} \,-\,4x^2(4x^2-1)^{-\frac{1}{2}}}{4x^2-1}\right]\)
Multiply top and bottom by (4x2−1)21
\(\displaystyle \L f'(x) \;= \;\frac{(4x^2-1)^{\frac{1}{2}}}{(4x^2-1)^{\frac{1}{2}}}\,\cdot\,4\left[\frac{(4x^2-1)^{\frac{1}{2}} \,- \,4x^2(4x^2-1)^{-\frac{1}{2}}}{4x^2-1}\right]\)
. . . . . . \(\displaystyle \L=\;4\left[\frac{4x^2\,-\,1\,-\,4x^2}{(4x^2-1)^{\frac{3}{2}}}\right] \;= \;\fbox{\frac{-4}{\left(4x^2\,-\,1\right)^{\frac{3}{2}}}}\)
[The "box" is for you, Cody!]
Galactus is absolutely correct . . .
Differentiate: \(\displaystyle f(x) \:=\:\L\frac{4x}{\sqrt{4x^2\,-\,1}}\)
Here's what I get:
\(\displaystyle \L f'(x) \;=\;\frac{{\left( {4x^2 - 1} \right)^{\frac{1}{2}}\cdot4\, - \,4x\cdot\frac{1}{2}\cdot\left(4x^2 - 1 \right)^{-\frac{1}{2}}}}{4x^2 - 1}
\\) ↖ times the derivative of 4x2−1
If there is a constant coefficient, I prefer to "leave it out front".
We have: \(\displaystyle \L\:f(x)\;=\;4\,\cdot\,\frac{x}{\left(4x^2-1\right)^{1/2}}\)
Then: \(\displaystyle \L f'(x) \;= \;4\,\left[\frac{(4x^2-1)^{\frac{1}{2}}\cdot1 \,- \,x\cdot\frac{1}{2}\cdot\left(4x^2-1\right)^{-\frac{1}{2}}\cdot8x}{4x^2-1} \right]\)
. . . . . . . . . . .\(\displaystyle \L=\;4\,\left[\frac{(4x^2-1)^{\frac{1}{2}} \,-\,4x^2(4x^2-1)^{-\frac{1}{2}}}{4x^2-1}\right]\)
Multiply top and bottom by (4x2−1)21
\(\displaystyle \L f'(x) \;= \;\frac{(4x^2-1)^{\frac{1}{2}}}{(4x^2-1)^{\frac{1}{2}}}\,\cdot\,4\left[\frac{(4x^2-1)^{\frac{1}{2}} \,- \,4x^2(4x^2-1)^{-\frac{1}{2}}}{4x^2-1}\right]\)
. . . . . . \(\displaystyle \L=\;4\left[\frac{4x^2\,-\,1\,-\,4x^2}{(4x^2-1)^{\frac{3}{2}}}\right] \;= \;\fbox{\frac{-4}{\left(4x^2\,-\,1\right)^{\frac{3}{2}}}}\)
[The "box" is for you, Cody!]