Finding derivative involving large powers.

• Jan 14th 2010, 07:02 PM
Calamanation
Finding derivative involving large powers.
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Hopefully one of those worked, just stuck on how to work question (2b) out, can't see a solution beyond something involving dominant powers. Thanks.
• Jan 14th 2010, 07:04 PM
pickslides
Which question in particular? What work have you done thus far?
• Jan 14th 2010, 09:36 PM
Calamanation
Question 2b involving calculating f'(x), I haven't been able to do much because all my research hasn't come across a question like that.
I appreciate it appears that I'm just looking for someone to answer it for me but studying for an exam has led me to asking for help.
• Jan 14th 2010, 09:46 PM
pickslides
It seems pretty straight forward, use the chain rule as the question is in the form $\displaystyle y= u^{1\over 2}$ where

$\displaystyle u= \frac{(\dots)^{12}(\dots)^{23}}{(\dots)^{10}}$ Just being lazy here!

Find $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\displaystyle \frac{du}{dx}$ can be found using the quotient rule

Does this help?
• Jan 14th 2010, 10:26 PM
Jose27
Apply $\displaystyle \ln$ on both sides and derive (remember $\displaystyle (\ln (y) )'=\frac{dy}{y}$ and your first exercise):

$\displaystyle \frac{f'(x)}{f(x)} = \frac{6(\cos (x) -\sin (x))}{4+\cos (x) + \sin (x)}$$\displaystyle + \frac{46x}{1+2x^2} -\frac{5(3x^2+4x^3)}{1+x^3+x^4}$

Now just isolate $\displaystyle f'$.
• Jan 15th 2010, 06:00 PM
Calamanation
Quote:

Originally Posted by pickslides
It seems pretty straight forward, use the chain rule as the question is in the form $\displaystyle y= u^{1\over 2}$ where

$\displaystyle u= \frac{(\dots)^{12}(\dots)^{23}}{(\dots)^{10}}$ Just being lazy here!

Find $\displaystyle \frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$

$\displaystyle \frac{du}{dx}$ can be found using the quotient rule

Does this help?

Helps heaps, thanks for that, very much appreciated
• Jan 15th 2010, 07:04 PM
Prove It
Quote:

Originally Posted by Calamanation
<a href="Math.jpg picture by Calamanation - Photobucket" target="_blank"><img src="http://i971.photobucket.com/albums/a...ation/Math.jpg" border="0" alt="Photobucket"></a>

http://i971.photobucket.com/albums/a...ation/Math.jpg

Hopefully one of those worked, just stuck on how to work question (2b) out, can't see a solution beyond something involving dominant powers. Thanks.

2. b) is probably easiest if you realise

$\displaystyle \sqrt{\frac{(4 + \cos{x} + \sin{x})^{12}(1 + 2x^2)^{23}}{(1 + x^3 + x^4)^{10}}} = \frac{(4 + \cos{x} + \sin{x})^6(1 + 2x^2)^{\frac{23}{2}}}{(1 + x^3 + x^4)^{5}}$.

Now you don't need to use chain rule inside chain rule.

2. c) $\displaystyle \frac{dy}{dx} = x^2 y$

$\displaystyle \frac{1}{y}\,\frac{dy}{dx} = x^2$

$\displaystyle \int{\frac{1}{y}\,\frac{dy}{dx}\,dx} = \int{x^2\,dx}$

$\displaystyle \int{\frac{1}{y}\,dy} = \frac{1}{3}x^3 + C_1$

$\displaystyle \ln{|y|} + C_2 = \frac{1}{3}x^3 + C_1$

$\displaystyle \ln{|y|} = \frac{1}{3}x^3 + C$, where $\displaystyle C = C_1 - C_2$

$\displaystyle |y| = e^{\frac{1}{3}x^3 + C}$

$\displaystyle |y| = e^Ce^{\frac{1}{3}x^3}$

$\displaystyle y = \pm e^C e^{\frac{1}{3}x^3}$

$\displaystyle y = Ae^{\frac{1}{3}x^3}$, where $\displaystyle A = \pm e^C$.

You are told $\displaystyle y(0) = 1$, so

$\displaystyle 1 = Ae^{\frac{1}{3}(0)^3}$

$\displaystyle 1 = Ae^0$

$\displaystyle 1 = A$.

So $\displaystyle y = e^{\frac{1}{3}x^3}$.

Can you do the rest of that question?
• Jan 15th 2010, 07:16 PM
Prove It
Also, to prove

$\displaystyle \ln{(XY)} = \ln{X} + \ln{Y}$

we look at the properties of exponentials.

We know that

$\displaystyle a^x\cdot a^y = a^{x + y}$.

So taking the logarithm of that base to both sides, we get

$\displaystyle \log_a{(a^x \cdot a^y)} = \log_a{(a^{x + y})}$

$\displaystyle \log_a{(a^x \cdot a^y)} = x + y$.

And since $\displaystyle x = \log_a{a^x}$ and $\displaystyle y = \log_a{a^y}$, we find

$\displaystyle \log_a{(a^x \cdot a^y)} = \log_a{a^x} + \log_a{a^y}$.

Using $\displaystyle e$ as your base and $\displaystyle X = a^x$ and $\displaystyle Y = a^y$ completes your proof.