Question

Differentiate:

Answer 1

\[\LARGE y=2^{cot~x}\]

Answer 2

Use logs
\[\Large \ln y = \cot x \ln 2\]Now use implicit differentiation

Answer 3

So the \[\LARGE lny\] becomes \[\LARGE \frac{1}{y'}\] right?
For the right I use product rule?

Answer 4

No the \(\bf ln(y)\) becomes:\[\bf \frac{ d }{ dx }\ln(y)=\frac{ 1 }{ y }*\frac{ dy }{ dx }\] This is a result of the chain rule.

Answer 5

For the right you have a constant (ln2) multiplied by cotx, no need for product rule.

Answer 6

Sorry, kind of re-learning everything again.

Answer 7

A somewhat easy way of remembering derivative of ln\[\LARGE \frac{ d }{ dx } \ln f(x) = \frac{ f'(x) }{ f(x) }\]

Answer 8

and
\[(\cot x)' = - \csc^2x = \frac{ -1 }{ \sin^2x } = -(1+\cot^2x)\]

Answer 9

\[\frac{ dy }{ dx }= 2^{cotx}.\ln2(-cosec^2x)\]

Answer 10

\[\huge y = 2^{\cot x}\]
\[\huge \ln y= \ln 2 *\cot x\]
\[\huge \frac{ 1 }{ y }y' = -\csc^2x\]
get y' by itself..
\[\huge y' = y(-\csc^2x)\]
replace y with the original equation
\[\huge y' = 2^{\cot x} * (-\csc^2x)\]

Answer 11

@shamil98 You forgot the ln(2) on the RHS

Answer 12

\[\huge (\ln(x))' = \frac{ 1 }{ x }\]

Answer 13

oh my bad ,
\[\huge y ' = 2^{\cot x} * (\ln 2 * -\csc^2x)\]

Answer 14

Alright, I see.. thanks guys!