What is the Derivative of (1+tanx)?

fitz_calc
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real quick review needed, I'm working through the quotient rule on a problem and need the first derivative of:

(1+tanx)

is it simply equal to:

(0 + sec^2x * 1)
(sec^2 x)

??
 
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yes that's correct

1 = 0, \tan{x}=\sec^2{x}
 
rocophysics said:
1 = 0
Geez, that's a nice identity.
I wish I could use that on my exam, it would make life a lot easier :biggrin:
 
CompuChip said:
Geez, that's a nice identity.
I wish I could use that on my exam, it would make life a lot easier :biggrin:
lol, you know what i meant ;)
 
Yeah, I know you meant

1' = 0, \tan(x)' = \sec^2(x)
where ' = \frac{\rm d}{{\rm d}x}.
 
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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