# A few questions on a HUGE derivatives assignment

1. Nov 27, 2011

### stonecoldgen

Hi physics forums, this post may seem like it has lots of questions, but don't think that you are doing ''my homework for me,'' it's actually a huge assignment, in which im only going to ask a few questions about.

this question is actualy easy (or seems to be), but i keep getting 64/3 and noon of the multiple choice answers

the slope of the line tangent to the curve y3+(xy-1)2=0 at the point
-9/4, -4 is:

A.0 B.1/3 C.32/59 D.8/3 E.16/3

I started by doing implicit differentiation getting
3y2dy/dx+2x2ydy/dx+2xy2-2xdy/dx-2y=0

then solved for dy/dx=(2y-2xy2)/(3y2+2x2y-2x)

then i just plugged in the numbers and got 64/3, any idea on what's wrong?

now this one:
let f be the function give by ef(x)=3x+1, g be the function given by 2lng(x)=4x2-6 and h be the function given by h(x)=f(g(x)). Find h'(x)

i rewrote both f and g the following way:
ln(3x+1)=f(x)
e2x2-3=g(x)

i insert g into f, but don't know how exactly would it lead be to one of the following options:

A.(12xe2x2-3)/(3e2x2-3+1)
B.(xe2x2-3)/(3e2x2-3+(1/3))
C.(3xe2x2-3)/(3e2x2-3+1)
D.(12e2x2-3)/(3e2x2-3+1)
E.(4xe2x2-3)/(3e2x2-3+1)

BTW, sorry if that was way too resumed up, it's just because of the hurry, hope that's enough

and THANKS A LOT, YOU GUYS ARE SAVING MY LIFE

EDIT: there is yet another question

suppose that the function f has a continous second derivative for all x, and that f(0)=2, f'(0)=-3 and f''(0)=0. Let g be a function whose derivative is given by g'(x)=e-2x(3f(x)+2f'(x)) for all x

a)write an equation of the line tangent to the graph of f at the point where x=0
b) given that g(0)=4, write an equation of the line tangent to the graph of g at the point where x=0
c)show that g''(x)=e-2x(-6f(x)-f'(x)+2f''(x))

WOAH, that question completely blows my mind.

I guess the equation for a) would be y=-3x+2
for b), it would be just plugging in numbers

in c) im completely lost

Last edited: Nov 27, 2011
2. Nov 27, 2011

### Dick

On the first one I think you got dy/dx right. You seem to have subbed the numbers wrong.

3. Nov 28, 2011

### HallsofIvy

Staff Emeritus
You might find it better to put x= -9/4, y= 4 here and then solve for dy/dx.

Don't. Instead, use the chain rule: h'(x)= f'(g(x))g'(x)

NEVER use the word "guess"- even when you do! Yes, that is correct.

Yes. Make it so!

You are given that $g'(x)= e^{-3x(3f(x)- 2f'(x))}$

Differentiate it!