1. ## Differentiate questions

Hi
Can someone help me differentiate the following questions:
1)$\displaystyle \frac{sinx+cosx}{sinx-cosx}$

2)Find the local maximum and minimum values of f(x) for each of the following and state the corresponding x value of each. (Consider $\displaystyle 0<= x <= 2\pi$only):
$\displaystyle f(x)=2cosx+sin2x$

P.S

2. Originally Posted by Paymemoney
Hi
Can someone help me differentiate the following questions:
1)$\displaystyle \frac{sinx+cosx}{sinx-cosx}$

2)Find the local maximum and minimum values of f(x) for each of the following and state the corresponding x value of each. (Consider $\displaystyle 0<= x <= 2\pi$only):
$\displaystyle f(x)=2cosx+sin2x$

P.S
First, this should be in the Calculus sub-forum, not Pre-calculus.

1) $\displaystyle y = \frac{\sin{x} + \cos{x}}{\sin{x} - \cos{x}}$.

To differentiate this, use the quotient rule.

$\displaystyle \frac{dy}{dx} = \frac{(\sin{x} - \cos{x})\frac{d}{dx}(\sin{x} + \cos{x}) - (\sin{x} + \cos{x})\frac{d}{dx}(\sin{x} - \cos{x})}{(\sin{x} - \cos{x})^2}$

$\displaystyle = \frac{(\sin{x} - \cos{x})(\cos{x} - \sin{x}) - (\sin{x} + \cos{x})(\cos{x} + \sin{x})}{(\sin{x} - \cos{x})^2}$

$\displaystyle = \frac{-(\sin{x} - \cos{x})^2 - (\sin{x} + \cos{x})^2}{(\sin{x} - \cos{x})^2}$

$\displaystyle = -1 - \left(\frac{\sin{x} + \cos{x}}{\sin{x} - \cos{x}}\right)^2$.

3. Originally Posted by Paymemoney
Hi
Can someone help me differentiate the following questions:
1)$\displaystyle \frac{sinx+cosx}{sinx-cosx}$

2)Find the local maximum and minimum values of f(x) for each of the following and state the corresponding x value of each. (Consider $\displaystyle 0<= x <= 2\pi$only):
$\displaystyle f(x)=2cosx+sin2x$

P.S
2) $\displaystyle f(x) = 2\cos{x} + \sin{(2x)}$

$\displaystyle f'(x) = -2\sin{x} + 2\cos{(2x)}$.

To find local maximums and minimums, let $\displaystyle f'(x) = 0$

$\displaystyle 0 = -2\sin{x} + 2\cos{(2x)}$

$\displaystyle 0 = -2\sin{x} + 1 - 2\sin^2{x}$

$\displaystyle 2\sin^2{x} + 2\sin{x} - 1 = 0$

Let $\displaystyle X = \sin{x}$, so

$\displaystyle 2X^2 + 2X - 1 = 0$

$\displaystyle X = \frac{-2 \pm \sqrt{2^2 - 4(2)(-1)}}{2(2)}$

$\displaystyle = \frac{-2 \pm \sqrt{12}}{4}$

$\displaystyle = \frac{-2 \pm 2\sqrt{3}}{4}$

$\displaystyle = \frac{-1 \pm \sqrt{3}}{2}$.

So $\displaystyle \sin{x} = \frac{-1 + \sqrt{3}}{2}$ or $\displaystyle \sin{x} = \frac{-1 - \sqrt{3}}{2}$.

Now get decimal approximations for $\displaystyle x$.

4. thanks for answering the questions.
I thought this is pre-uni maths.

5. It's clearly calculus. Therefore it goes in the calculus sub-forum.

It's a bit misleading I know, but there you go. You should have also read the "Read this before posting" thread, where it is explained.