1. f(x) = (4x +5)^x. Do the following.

a.) Find dy/dx. Hint: Use logarithmic differentiation, but do not simplify the derivative.

my work:

y= (4x+5)^x
lny = xln(4x+5)
(1/y) dy/dx = x * (1/(4x+5))
(1/y) dy/dx = x/(4x+5)
dy/dx = xy/(4x+5) <------------ answer (i am VERY sure of my answer)


b.) Find the slope of the line tangent to the curve when x=-1

my work:
f(x) = (4x+5)^x
f'(x) = 4x(4x+5)^(x-1)
4(-1)(4(-1) +5)^(-1-1)
-4(-4 +5)^-2
-4(1)^-2
-4*1
-4 <----------------------- answer (i am skeptical about my answer and feel that something was done wrong)

c) Find the value of the limit of [(4x+5)^x - 1]/(x+1) when x approaches -1. Explain your process

Not sure on this whatsoever!

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Actually i forgot there was a backside :x. If anyone can help me on these last two that would be greatly appreciated :).


A particle moves on a vertical line so that its coordinate at time t is y=t^3 - 3t - 15 for t>0 where y is measured in centimeters and t is in seconds and its velocity is v = 3t^2 - 3

a.) Determine when the particle is moving upward. Explain.

My answer:
t>1
The particle is moving upward when t>1, or in other words after 1 second has elapsed. Before 1 second it is moving downwards, at 1 second it is not moving at all. When "v" is positive in v =3t^2 - 3, it is moving upwards.


b.) Find the distance that the particle travels in the time interval 0< t < 3

Not sure how to go about this one.

This post was edited by CamelFinger on Nov 6 2013 08:14pm

1c what does x approach? or is that all the information youre given

im pretty sure part a is right. might want to include that when you have y(0) the answer is "negative cm" or it has moved down the vertical line

if you plug in 0 for t, the y value will be how many cm the particle has moved

similarly, if you plug in 3 for t, the y value reflects how many cm the particle has moved in 3 seconds

so the total distance it travels will be [y(3) - y(0)]

This post was edited by nlin on Nov 6 2013 08:12pm

1c can be solved with lhopitals rule if youve learned it

When using the power rule for derivative, the power must be a constant.

Try, instead, to use the chain rule :

y(x) =(4x + 5)^x = exp(x.ln(4x+5))

y' (x) = (x.ln(4x+5))'.y(x) = [ (x)'.ln(4x+5) + x.(ln(4x+5))' ] . y(x)

So you got wrong on a) ...

For c), just observe that the limit you're asked for is the limit of : (y(x) - y(-1)) / ( x - (-1)) as x approaches -1.
It is the exact definition of y'(-1). So the answer for c) is the exact slope you already found at b ) : -4.

For the second problem, the particule is moving upward as long as its velocity is positive. Just solve for t : 3t² - 3 > 0 ( answer t > 1 is correct, given that t>=0 ).

Now for the distance that the particule travels between t=0 and t=3...
Its initial position is y(0) and its final position is y(3), so the difference is [y(3) - y(0)] as nlin said.
But... since the particule moves downward between t=0 and t=1, you can also understand the problem as : |y(1) - y(0)| + [y(3) - y(1)]
(and I would interpret like that honestly)
That is to say, the distance traveled in a way shouldn't be canceled by the same distance traveled in the other way (going down, then up).

This post was edited by feanur on Nov 7 2013 12:07am

Can anybody clarify this more?

I am still lost

Need to know in a few hours

not sure what they mean by do not simplify the derivative, but here goes a more explicit version of what feanur did:

I'll give you a hint, it is going to be an integral...