find y'(pi/4) if y(x)=(6sinx)^x y'(pi/4)=
does y'= 6^x sin^x(x) (x cot(x)+log(6 sin(x)))
since the exponent is a variable then I would use logarithm differentiation
ok so the x goes to the front
xln(6sinx)
So you have something in this form: \[y=f^g \] => \[\ln(y)=g \ln(f) \text{ for } f>0\] Then differentiate both sides like so: \[\frac{y'}{y}=g' \cdot \ln(f)+g \cdot \frac{f'}{f} \\ \text{ applied product rule here along with some other rules of course }\]
Then you solve that for y' by multiplying y on both sides giving you: \[y'=y(g' \ln(f)+g \frac{f'}{f})\]
after that where does the pi/4 go?
after you find y' you replace x with pi/4
because y'(pi/4) tells us to do that
y'(pi/4) means to find y' first and then to take y' and evaluate it for x=pi/4
ok ill finish working through it
ok so y'(pi/4)=ln(6sin(x))+x(6cos(x))/(6sin(x))
idk i probably screwed it up
that is't y'(pi/4) y'(pi/4) will be a constant looks like you found y'/y
ok what next
recall your objective is to find y'(pi/4) which means you need to find y'
first
6^x sin^x(x) (x cot(x)+log(6 sin(x)))
\[\frac{y'}{y}=\ln(6 \sin(x))+\frac{x \cos(x)}{\sin(x)}\] to solve for y' you multiply y on both sides \[y'(x)=(6\sin(x))^x(x \cot(x)+\ln(6 \sin(x)))\]
which is what you wrote if you mean that log to be natural log
now plug in pi/4
Don't be nervous about your answer if it doesn't turn out pretty
unless you are able to approximate
what i'm saying is the exact answer is not pretty at all
and it can't be simplify to a more pretty form
of course you can simplify some parts but in all what i mean is you won't get it to look much prettier than it already is which is pretty ugly
yeah i understand. this answer is cringe worthy 2^(pi/8) 3^(pi/4) (pi/4+log(3 sqrt(2)))
2^(pi/8) 3^(pi/4) (pi/4+log(3 sqrt(2)))
can i get a head start with this problem. i want to work through this on my own. what's troubling me is x such that (0,pi/2)
thanks a lot btw
no problem
i will look at your screenshot and try to untrouble you
Oh. I think they restricted the domain of the function because recall tan is not defined everywhere
This problem is same as last let y=x^(2tan(x)) take the same process we did before
Well it isn't a thinking thing. It is exactly what they did, not thinking, but pure knowing. lol
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