Find the derivative using Logaritmic Differentiation

[superjen]

y = (sinx)^2x

LNy = 2xLN(sinx) + (1 over sinx)(cosx)(2x)

Answer

y' = (sinx)^2x [2cosx over sinx + 2xcotx]


and

y = (cosx)^cosx

i did it the same way as above
Answer i got was

y' = (cosx)^cosx [ (-sinx)(cosx) + sinx]

am i anywhere right?

 

[wywong]

y = (sinx)^2x

LNy = 2xLN(sinx) + (1 over sinx)(cosx)(2x)

How comes the right term??

$$ln(y) = 2xln(sinx)$$
$$~~\frac{y'}{y} = 2(ln(sinx)+xcotx)$$
$$~~y' = 2sinx^2x(ln(sinx)+xcotx)$$

y' = (cosx)^cosx [ (-sinx)(cosx) + sinx]
am i anywhere right?

Nope. Please try again. There should be a ln() in your answer.

 

[superjen]

for the second one , would this be right?

y' = (cosx)^cosx[(-sinx)(LN(sinx) - sinx]

 

[Dick]

for the second one , would this be right?

y' = (cosx)^cosx[(-sinx)(LN(sinx) - sinx]

Yes, it would.

 

[wywong]

for the second one , would this be right?

y' = (cosx)^cosx[(-sinx)(LN(sinx) - sinx]

I don't think so. I think it should be ln(cosx) instead of ln(sinx). Typo?