Finding partial derivative of a trig function

lagwagon555
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Homework Statement



Find the partial derivative with respect to x of sin(xyz - 1)


Homework Equations



None needed.

The Attempt at a Solution



I took the answer to be yz*cos(xyz - 1), but wolfram alpha is giving me yz*cos(1 - xyz). Anyone know what's going on here? Thanks!
 
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cos(x) is equal to cos(-x)

Those answers are equivalent.

Btw, if you're confused as to what Wolfram Alpha is doing, press the "Show steps" button. Helps out sometimes.
 
Ah right I see, thanks a lot for clarifying that. I'm taking a senior level maths paper after 2 years of not doing maths, so I keep slipping up on things like that. Thanks again :)
 

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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