A few questions on a HUGE derivatives assignment

  • #1
stonecoldgen
109
0
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:

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
620
On the first one I think you got dy/dx right. You seem to have subbed the numbers wrong.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
43,021
970
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
You might find it better to put x= -9/4, y= 4 here and then solve for dy/dx.

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

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

for b), it would be just plugging in numbers
Yes. Make it so!

in c) im completely lost
You are given that [itex]g'(x)= e^{-3x(3f(x)- 2f'(x))}[/itex]

Differentiate it!
 

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