Trifunction Derivatives problem

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I have questions on derivatives cause today I did not go to school and miss Caculus AB and we went over Trifunction Derivatives.

So can anybody show me the rules or send me a link cause I would ask a friend the notes but sometimes my teacher gives bad notes...
 
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cracker said:
I have questions on derivatives cause today I did not go to school and miss Caculus AB and we went over Trifunction Derivatives.

So can anybody show me the rules or send me a link cause I would ask a friend the notes but sometimes my teacher gives bad notes...

Do you mean the derivative rules for trigonometric functions?
http://www.sosmath.com/calculus/diff/der03/der03.html
 
The quotient Rule is:
y= u/v

y'= [(vu'-uv')/(v^2)]

The product Rule is:
y= uv

y'= (vu'+uv')

the chain Rule is:

y=u^n

y'=(n)(u^n-1)(u')
 
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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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