# Math Help - Finding derivative involving large powers.

1. ## 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.

2. Which question in particular? What work have you done thus far?

3. 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.

4. 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?

5. 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 $f'$.

6. 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

7. Originally Posted by Calamanation
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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.
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?

8. 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.