# Thread: Having Lots Of Calculus III Problems

1. ## Having Lots Of Calculus III Problems

My brain has gone dead, got a few days to do the following questions:

would appreciate any help with any of the above question and methods

thank you

2. Originally Posted by amanda1410
My brain has gone dead, got a few days to do the following questions:

would appreciate any help with any of the above question and methods

thank you
Q4a. Let:

$
f(x)=x^3-2x^2+x-3
$

then $f(2)=-1$ and $f(3)=9$. Also as $f(x)$ is continuous by the intermediate value theorem it takes all values between $f(2)$ and $f(3)$ between $x=2$ and $x=3$, but as $f(x)$ changes sign in this interval one of these values is $0$. Hence we conclude that $f(x)$ has a root in $(2,3)$.

RonL

3. cheers for that, appreciate all the help possbile with the rest, at uni pullin my hair out trying to do this and my electronics at the same time

4. First I would like to mention that $f''_{xy}$ and $f''_{yx}$ are the same! Because the function is continous on some open disk.
(But since you are in mechanical engineering you probably do not care about that).

$f(x,y)=x\ln y+e^{xy}+\cos (2x-3y)$
Thus,
$f'_x(x,y)=\ln y+ye^{xy}-2\sin (2x-3y)$
$f'_y(x,y)=\frac{x}{y}+xe^{xy}+3\sin (2x-3y)$

Now take the derivative along y on the second one,
$f''_{yy}(x,y)=-\frac{x}{y^2}+x^2e^{xy}-9\cos (2x-3y)$

And finally,
$f''_{xy}=\frac{1}{y}+e^{xy}+xye^{xy}+6\cos (2x-3y)$

5. Question 1 a) I presume.

i) $y = (2x - 1)3^{5x+1} - tan(2x)$

First note that $\frac{d}{dx}a^x = ln(a) \cdot a^x$. The rest is an exercise in the chain rule:

So
$y' = 2 \cdot 3^{5x+1} + (2x - 1) \cdot ln(3) \cdot 3^{5x+1} \cdot 5 - sec^2(2x) \cdot 2$

$y' = [ 2 + 5(2x - 1)ln(3) ] 3^{5x+1} - 2 sec^2(2x)$

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ii) $y = ln \left ( 2x + e^{2x} \right ) + sin( \pi x)$

$y' = \frac{1}{2x + e^{2x}} \cdot (2 + e^{2x} \cdot 2 ) + cos( \pi x) \cdot \pi$

$y' = \frac{2(1 + e^{2x})}{2x + e^{2x}} + \pi cos( \pi x)$

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iii) $y = x + cos^{-1}(y)$

First, this needs to be done by implicit differentiation.
Second: $\frac{d}{dx}cos^{-1}(x) = - \frac{1}{\sqrt{1-x^2}}$.

So:
$y' = 1 + -\frac{1}{\sqrt{1-y^2}} \cdot y'$

$y' + \frac{1}{\sqrt{1-y^2}} \cdot y' = 1$

$y' \left ( 1 + \frac{1}{\sqrt{1-y^2}} \right ) = 1$

$y' = \frac{1}{1 + \frac{1}{\sqrt{1-y^2}}}$

This needs to be neated up a bit:
$y' = \frac{1}{1 + \frac{1}{\sqrt{1-y^2}}} \cdot \frac{\sqrt{1-y^2}}{\sqrt{1-y^2}}$

$y' = \frac{\sqrt{1-y^2}}{\sqrt{1-y^2} + 1}$

$y' = \frac{\sqrt{1-y^2}}{\sqrt{1-y^2} + 1} \cdot \frac{\sqrt{1-y^2} - 1}{\sqrt{1-y^2} - 1}$

$y' = \frac{(\sqrt{1-y^2})(\sqrt{1-y^2} - 1)}{1 - y^2 - 1}$

$y' = -\frac{1 - y^2 - \sqrt{1-y^2}}{y^2}$

I'll leave it like this.

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iv) Ah, a practically straightforward problem!
$y = \frac{\sqrt{2x+1}}{x}$

Using the quotient rule:
$y' = \frac{\left ( \frac{1}{2} \frac{1}{\sqrt{2x+1}} \cdot 2 \right ) x - \sqrt{2x+1} \cdot 1}{x^2}$

$y' = \frac{\left ( \frac{1}{\sqrt{2x+1}} \right ) x - \sqrt{2x+1}}{x^2}$

$y' = \frac{\left ( \frac{1}{\sqrt{2x+1}} \right ) x - \sqrt{2x+1}}{x^2} \cdot \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$

$y' = \frac{x - (2x+1)}{x^2 \sqrt{2x+1}}$

$y' = \frac{-x - 1}{x^2 \sqrt{2x+1}} \cdot \frac{\sqrt{2x+1}}{\sqrt{2x+1}}$

$y' = -\frac{(x+1)\sqrt{2x+1}}{x^2 (2x+1)}$

-Dan