# Finding derivative involving large powers.

• January 14th 2010, 07:02 PM
Calamanation
Finding derivative involving large powers.
<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.
• January 14th 2010, 07:04 PM
pickslides
Which question in particular? What work have you done thus far?
• January 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.
• January 14th 2010, 09:46 PM
pickslides
It seems pretty straight forward, use the chain rule as the question is in the form $y= u^{1\over 2}$ where

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

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

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

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

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

Now just isolate $f'$.
• January 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 $y= u^{1\over 2}$ where

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

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

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

Does this help?

Helps heaps, thanks for that, very much appreciated
• January 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$1 = Ae^0$

$1 = A$.

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

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

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

we look at the properties of exponentials.

We know that

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

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

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

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

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

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

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