derivatives

[jas14]

find tbe derivative of the following

tan^2(x)

cosx^2 / 3x

sin^-1(2x)

ln(x^2 + 2x)

4^(3x-1)^2

thanks

 

[pka]

See my advise at http://www.mathgoodies.com/forums/topic.asp?TOPIC_ID=26786

 

[soroban]

Hello, jas14!

Sorry, but I have to ask: Don't you know any of the differentiation rules ??
. . If you're like 99% of the students, your response is "I just wanted to check my answers."
. . If that's true, why not show us your answers?

Having said all of that, here are a few of them . . .

\(\displaystyle y\;=\;\tan^2(x)\)

Chain Rule: .\(\displaystyle y'\;=\;2\cdot\tan(x)\cdot\sec^2(x)\)

cosx^2/3x

Is that: .\(\displaystyle \frac{\cos(x^2)}{3x}\) or \(\displaystyle \frac{\cos^2(x)}{3x}\) ?

\(\displaystyle y\;=\;\sin^{-1}(2x)\)

\(\displaystyle y'\;=\;\frac{1}{\sqrt{1\,-\,(2x)^2}}\,\cdot\,2 \;=\;\frac{2}{\sqrt{1\,-\,4x^2}}\)

\(\displaystyle y\;=\;\ln(x^2 + 2x)\)

\(\displaystyle y'\;=\;\frac{1}{x^2\,+\,2x}\cdot(2x + 2) \;=\;\frac{2(x\,+\,1)}{x(x\,+\,2)}\)

4^(3x-1)^2

You need more parentheses . . .

It could be: \(\displaystyle 4^{(3x-1)^2}\;=\;4^{(9x^2-6x+1)}\)

. . . . or: .\(\displaystyle [4^{(3x-1)}]^2\;=\;4^{(3x-1)\cdot2} \;=\;4^{(6x-2)}\)

 

[happy]

Nice advice on the "other site", pka! :wink: