- #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 I am 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 continuous 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) I am completely lost
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 continuous 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) I am completely lost
Last edited: