1. ## Derivative

Evaluate the derivative of the function below at the point (, ).

y=7/x + sqrt cos(x)

2. well the derivative of that function is just
-7/(x^2)+[1/(2(sqrt(cos(x)))]*-sin(x)

evaluation of derivative of sqrt[cos(x)] is just applying the chain rule.
now all u have to do is substitute the value of x in.

3. This is the part I don't get...actually putting TT/4 into the equation...

I know cos(x) = sqrt2/2 and the same for sin (x), right?

Evaluate the derivative of the function below at the point (, ).

y=7/x + sqrt cos(x)
First, rewrite the function like this:
$y = 7x^{-1} + (\cos{x})^{\frac{1}{2}}$

then, by the chain rule:

$y^{\prime} = -7x^{-2} + \frac{1}{2}(cos{x})^{-\frac{1}{2}}(-\sin{x})$

and $y^{\prime}\left(\frac{\pi}{4}\right) = -11.76844$:
• $\frac{\pi}{4}$ is not a critical number
• the curve is falling at $\left(\frac{\pi}{4}, 9.754\right)$

Evaluate the derivative of the function below at the point (, ).

y=7/x + sqrt cos(x)
$\frac{dy}{dx}=\frac{-7}{x^2}-\frac{1}{2}(cosx)^{-\frac{1}{2}}.sinx$

$\frac{dy}{dx}=\frac{-7}{(\frac{\pi}{4})^2}-\frac{1}{2}(cos(\frac{\pi}{4}))^{-\frac{1}{2}}.sin(\frac{\pi}{4})$ $=-11.77$

This is the part I don't get...actually putting TT/4 into the equation...

I know cos(x) = sqrt2/2 and the same for sin (x), right?
Right. Just to clarify:

$
y^{\prime} = -7x^{-2} + \frac{1}{2}(cos{\frac{\pi}{4}})^{-\frac{1}{2}}(-\sin{\frac{\pi}{4}}) = -7\left(\frac{\pi}{4}\right)^{-2} + \frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)^{-\frac{1}{2}}\left(-\frac{\sqrt{2}}{2}\right) = -11.7684
$

7. Originally Posted by sinewave85
Right. Just to clarify:

$
y^{\prime} = -7\color{red}{x}\color{black}^{-2} + \frac{1}{2}(cos{\frac{\pi}{4}})^{-\frac{1}{2}}(-\sin{\frac{\pi}{4}}) = -7\color{red}{x}\color{black}^{-2} + \frac{1}{2}\left(\frac{\sqrt{2}}{2}\right)^{-\frac{1}{2}}\left(-\frac{\sqrt{2}}{2}\right) = -0.4408563709
$
I think there is a mistake!

8. Originally Posted by u2_wa
I think there is a mistake!
Big oops. Thanks. I somehow lost that x as I went along.