- #1

stonecoldgen

- 109

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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 y

-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

3y

then solved for dy/dx=(2y-2xy

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 e

i rewrote both f and g the following way:

ln(3x+1)=f(x)

e

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

A.(12xe

B.(xe

C.(3xe

D.(12e

E.(4xe

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

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

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

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 y

^{3}+(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

3y

^{2}dy/dx+2x^{2}ydy/dx+2xy^{2}-2xdy/dx-2y=0then solved for dy/dx=(2y-2xy

^{2})/(3y^{2}+2x^{2}y-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 e

^{f(x)}=3x+1, g be the function given by 2lng(x)=4x^{2}-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)

e

^{2x2-3}=g(x)i insert g into f, but don't know how exactly would it lead be to one of the following options:

A.(12xe

^{2x2-3})/(3e^{2x2-3}+1)B.(xe

^{2x2-3})/(3e^{2x2-3}+(1/3))C.(3xe

^{2x2-3})/(3e^{2x2-3}+1)D.(12e

^{2x2-3})/(3e^{2x2-3}+1)E.(4xe

^{2x2-3})/(3e^{2x2-3}+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 xa)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

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