Derivative of a function, Is my answer correct?

neutron star
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Homework Statement


y=e^9Θsin(8Θ)


Homework Equations





The Attempt at a Solution


dy/dΘ=e^9Θ(sin(8Θ)+8cos(8Θ))
 
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[Edit] I didn't catch the missing 9; sorry about that.
 
Last edited:
is that (using t for theta)

y=e^(9t.sin(8t)) or y=(e^(9t)).sin(8t)?
 
Ugh, it said incorrect...
 
if it is
y=(e^(9t)).sin(8t)

i think you should get an extra factor of 9
y'=(e^(9t)).(9sin(8t)+ 8cos(8t))
 
Lanedance is right, you get that factor of 9 when taking the exponential derivative.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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