# Math Help - Calc Derivative Questions help plz

1. ## Calc Derivative Questions help plz

If anyone could lend some help on these that would be great!

Find the derivative:

1. f(x)=sinx-2tanx

2. y=tan(sinx^2)

3. f(x)= tan^-1 sq.root(1-x)

4. y=(secx+tanx)

5. Let f(x) = 7sin(x+pi)+cos2x
a. compute f'(x), f''(x), f'''(x), f''''(x)
b. compute f*13th derivative*(0)

2. Originally Posted by Dr. Noobles
If anyone could lend some help on these that would be great!

Find the derivative:

1. f(x)=sinx-2tanx
Derivative of sine is cosine.
Derivative of tangent is,
$\sec^2 x$
Thus,
$f'(x)=\cos x-2\sec^2x$
2. y=tan(sinx^2)
This is a double chain rule,
$y=\tan v$
$v=\sin u$
$u=x^2$
Thus,
$\frac{dy}{dv}=\sec^2 v$

$\frac{dv}{du}=\cos u$

$\frac{du}{dx}=2x$
Thus,
$\frac{dy}{dx}=\frac{dy}{dv}\cdot \frac{dv}{du}\cdot \frac{du}{dx}$
Thus,
$\frac{dy}{dx}=\sec^2 v\cos u (2x)$
Thus, substitute,
$\frac{dy}{dx}=2x\sec^2 (\sin u) \cos x^2$
Thus, subsitute again,
$\frac{dy}{dx}=2x\sec^2 (\sin x^2)\cos x^2$.

3. thanks ThePerfectHacker

Anyone have an idea on the others?

4. Originally Posted by Dr. Noobles
If anyone could lend some help on these ...
Find the derivative:
...
3. f(x)= tan^-1 sq.root(1-x)

4. y=(secx+tanx)
...
Hello,

to 3. You have to use the chain rule many times:

$\frac{df}{dx}=(-1)\cdot \frac{1}{\tan^2(\sqrt{1-x})} \cdot \frac{1}{\cos^2(\sqrt{1-x})} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{1-x}} \cdot (-1)$
Split the tan^2 into a fraction of (sin^2)/(cos^2) and cancel the cos^2. You'll get:
$\frac{df}{dx}=\frac{1}{2\sqrt{1-x} \cdot \sin^2(\sqrt{1-x})}$

to 4. I'll use: $y=(\sec(x)+\tan(x))=\frac{1}{\cos(x)}+\tan(x)$

$\frac{df}{dx}=(-1)\cdot \frac{1}{\cos^2(x)} \cdot (-\sin(x))+\frac{1}{\cos^2(x)}$
Simplify to:
$\frac{df}{dx}=\frac{1+\sin(x)}{\cos^2(x)}$

EB

5. Originally Posted by Dr. Noobles
3. f(x)= tan^-1 sq.root(1-x)
Is this
$f(x)= \frac{1}{tan \left ( \sqrt{1-x} \right ) }$

or

$f(x) = tan^{-1} \left (\sqrt{1-x} \right )$ where $tan^{-1}$ is the inverse tangent function.

Earboth did the first one, so in case you need it, I'll do the other.

$f(x) = tan^{-1} \left (\sqrt{1-x} \right )$

This is done by the chain rule just like the others. I'll note that:
$\frac{d}{dx}tan^{-1}(x) = \frac{1}{x^2+1}$

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

$f'(x) = - \frac{1}{2 (2 - x) \sqrt{1-x} }$

-Dan