# Homework Help: Calculus Derivative Question

1. Mar 28, 2005

### ms. confused

Hey,

Anyone out there able to help me out? I'm trying to find y'' of the equation y=xtanx. I found y' to equal 1(sec²x) but I don't know what to do after that. I know the final answer should be 2cosx + 2xsinx/ cos³x after being simplified and stuff, but I am clueless as to how to get there. Thanks in advance to anyone who can help me!

2. Mar 28, 2005

### dextercioby

Nope,it can't be.Are u sure u want to compute

$$\frac{d^{2}(x\tan x)}{dx^{2}}$$...?

Then better apply the Leibniz rule with care.

Daniel.

3. Mar 28, 2005

### ms. confused

You mean 2cosx + 2xsinx/ cos³x isn't the right answer or the y' I came up with to help me get the y'' isn't right?

4. Mar 28, 2005

### dextercioby

Do you how to get to this result

$$\frac{d(x\tan x)}{dx}=\tan x +x\sec^{2}x$$...?

If u do,i don't see any reason not to compute the 2-nd derivative correctly.

Daniel.

5. Mar 28, 2005

### dextercioby

Oh,your final aswer for the 2-nd derivative is in terms of "sin" & "cos",u'd better conver the 1-st derivative to a form containing "sin" a "cos" b4 the differentiation.

Daniel.

6. Mar 28, 2005

### Data

Your first derivative is wrong. You need to use the product rule. Do you remember it?

7. Mar 28, 2005

### tutor69

Data is right use the product rule to find first derivative and in finding the second derivative also.

$$\frac{d(x\tan x)}{dx}=\tan x +x\sec^{2}x$$

$$\frac{d^2(x\tan x)}{dx^2}= \frac{(2)(\cos{x} +x\sin{x})}{\cos^3{x}}$$

Last edited: Mar 28, 2005
8. Mar 28, 2005

### Data

indeed. Or for the second derivative you can just take it without converting to sines and cosines and get $2\sec^2{x} + 2x\sec^2{x}\tan{x}$, an equivalent answer.

9. Mar 29, 2005

### ms. confused

Where did you get the 2's from? Why isn't it just $sec^2{x} + sec^2{x}\tan{x}$

10. Mar 29, 2005

### Data

$$\frac{d}{dx} (\tan{x} + x\sec^2{x}) = \sec^2{x} + \frac{d}{dx} x\sec^2{x}$$

$$= \sec^2{x} + \left(\sec^2{x}\frac{d}{dx} x + x \frac{d}{dx}\sec^2{x}\right)$$ <------- Product Rule

$$= \sec^2{x} + \left(\sec^2{x} + x\left(2(\sec{x}\tan{x})\sec{x}\right)\right)$$ <----- Chain Rule

$$= \sec^2{x} + \sec^2{x} + 2x\sec^2{x}\tan{x}$$

$$= 2\sec^2{x} + 2x\sec^2{x}\tan{x}$$

Remember, when you're taking the derivative of a product, you need to use the product rule:

$$\frac{d}{dx} f(x)g(x) = f^\prime(x)g(x) + g^\prime(x)f(x) \neq f^\prime(x)g^\prime(x)$$

Last edited: Mar 29, 2005