# Differentiation problems

1. Jul 27, 2009

### JakePearson

have a few problems with these questions, can you help :)

1. f(x) = -x2 + x
2. f(x) = lnx - x
3. f(x) = -x4 + 2x2
4. f(x) = x2/4 + 4/x
5. f(x) = xe-2x2
6. f(x) = sqrt(x - n)/x ; n>0

hope you guys can help !

2. Jul 27, 2009

### Cyosis

Can you help as in do your homework for you? Certainly not! Before we help you we require you to show some work. You're not going to tell me that you have no clue whatsoever about 1) for example. If you're following a text you can be sure that there are examples on how to differentiate polynomials. Show us your attempts so far.

3. Jul 27, 2009

### HallsofIvy

Staff Emeritus
$dx^n/d= nx^{n-1}$
$d(ln(x))/dx= 1/x$
$de^x/dx= e^x$
d(uv)/dx= u dv/dx+ du/dx v
d(u/v)/dx= (du/dx v- u dv/dx)/v2

Surely, those are all in your textbook?

4. Jul 27, 2009

### JakePearson

hey guys, i do apologise, i had a go at them, and i was wondering if the answers i got were correct!

1.f(x) = -x^2 + x

d/dx(x^n) = nx^(n-1)

f '(x) = -2x + 1

2.f(x) = lnx - x

f '(x) = (1/x) - 1

3.f(x) = -x^4 + 2x^2

f '(x) = -4x^3 + 4x

4.f(x) = x^2 / 4 + 4 / x

For (x^2/4), pull out (1/4) and use the power rule. for 4/x use quotient rule.

d/dx(f(x)/g(x)) = (f '(x)g(x) - f(x)g '(x))/( g(x) ^2)

f '(x) = (1/2)x - 4/x^2

5. f(x) = xe^-2x^2

Product rule: d/dx(f(x)g(x)) = f'(x)g(x) + f(x)g '(x)
Chain rule: d/dx(f(g(x)) = f '(g(x)) * g'(x)

f '(x) = e^-2 * x^2 + x(2e^-2x)

6. f(x) = sqrt(x - n) / x ; n>0

sqrt(x - n) = (x - n)^ (1/2)

f '(x) = (1/2) (x - n)^(-1/2)

i do apologise again, hope u can help :)

5. Jul 27, 2009

### Chewy0087

1-4 are fine, 5 however;

f(x) = x * e^-2x^2

is more tricky, my advice here is first taking the natural log of both sides otherwise it gets very messy, you've missed alot out in your answer which is understandable, here i'll get you going;

ln (f(x)) = ln (x) + (-2x^2 * ln (e))

follow this rigidly, yes it'll be hard but it's a really good tool!

6;

You've ignored the /x

6. Jul 27, 2009

### Staff: Mentor

For #4 you got the right answer, but using the quotient rule on 4/x is overkill. If you rewrite 4/x as 4x-1 you can use the constant multiple rule and the power rule.

I.e., d/dx(4/x) = d/dx(4x-1) = 4*d/dx(x-1) = -4x-2

You should never use the quotient rule when either the numerator or denominator of the expression to be differentiated is a constant. The quotient rule will work, but there is a greater likelihood of making a simple algebraic error and thereby getting the wrong answer.

7. Jul 28, 2009

### nickmai123

For 5: The best thing to do is product rule. Do you know the rule for e^f(x)?

$$\frac{d}{dx} [e^{u}] = e^{u}u'$$

I learned it as, "deriv of e^whatever = e^whatever * deriv-whatever."