Using the Chain Rule to Find the Derivative of a Complex Function: Homework Help

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Homework Statement



lim
x-o+ (tan(4x))^x

Homework Equations





The Attempt at a Solution



to get the derivative i have to use the chain rule so it would be.

lim
x- 0+ (x(tan(4x)^x-1)(sec^2(4)
 
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Hey, jpd5184.

First you have to identify what kind of indeterminate form you have. Can you see you have the form 0^0?

To apply L'Hospital's rule, remember that you need to have \frac{0}{0} or \frac{\pm \infty}{\pm \infty}. What can do you do to transform this 0^0 indeterminate form to one of these? Hint: think of the log function and its properties.

Good luck.
 
Also, although you do not need it here because you do NOT just differentiate the function itself, the derivative, with respect to x, of f(x)^x is NOT "x f(x)^{x-1}". The power rule only works when the power is a constant, not a function of x.
 
would i just make tanx into sinx/cosx
 
I don't think that would be useful to do.

For this type of problem there is a technique that is useful.
Let y = (tan(4x))x
Then ln y = x ln(tan(4x)) = tan(4x)/(1/x)

Now take the limit of both sides, and recognize that what you're getting is the limit of the ln of what you want.

Check your textbook. I'm betting that there is an example that uses this technique.
 
thanks very much, i did learn this technique, just forgot it.
 
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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