Question

How do you find the slope of the following functions at the given points?:
f(x)=-1/2+7/5x^3 at the point (3/5, 2)
y=(2x+1)^2 at the point (0,1)
f(x)=3(5-x)^2 at the point (5,0)
f(theta)=4sin(theta)-(theta) at the point (0,0)
g(t)=2+3cos(t) at the point (pi, -1)

Answer 1

Find first derivative, then plug in the x (or theta or t) value from the point given.

Answer 2

I'm having trouble finding the derivative for some for these. I'll work out one of the problems and tell me what I did wrong.

f(x)=3(5-x)^2 for the point (5,0)
f(x)=3(25-10x+x^2)
f(x)=75-30x+3x^2
f'(x)=75-30(x^0)+3(2x)
f'(x)=75-30+6x
f'(5)=30

The answer key and the calculator says that the answer is 0 instead of 30. Where did I go wrong?

Answer 3

When you take the derivative of a constant, e.g. that 75 there, it becomes zero.

Answer 4

You can also differentiate via chain rule so you don't have to simplify the expression first.

Answer 5

\[f(x)=3(5-x)^2 \rightarrow f'(x)=-1 \cdot 2 \cdot 3 \cdot (5-x)^1 \rightarrow -6(5-x)\]
At x=5, f'(5)=0.

Answer 6

I got lost at the second step. Can you tell me what you did?

Answer 7

Sure, the chain rule says that when you differentiate a function of a function ((5-x)^2 is a function of 5-x, which itself is a function of x.) you multiply the derivative of the inside function by the derivative of the outside function.

Answer 8

For 3(5-x)^2, first multiply by the derivative of the inside function (the derivative of 5-x is -1), then take the derivative of the squaring function which is 2.

Answer 9

Altogether, f'(x) = -1*2*3*(5-x) = -6(5-x). Don't bother simplifying to 6x-30. Plugging 5 into (5-x) makes it obvious that the answer is zero.

Answer 10

Is there another one you're having trouble with?

Answer 11

Yes. The first problem I posted and one of the trig problems. (Thanks for the explanation on the third problem. I think I understand it now!)

Answer 12

Show me what you did for the first one and I'll try the f(theta) one.

Answer 13

The derivative of 4sin(Θ) - Θ = 4cos(Θ) - 1.
At Θ=0, f' = 3.

Answer 14

Okay, so this is what I did:

f(x)=-1/2+7/5x^3 at the point (0, -1/2) [sorry, I wrote the point down wrong in my question]
f(x)=-1/2+7/5(3x^2)
f'(x)=-1/2+(21/5)x^2
f'(0)=0

Is this the right way to solve this one?

Answer 15

Everything is fine except remember that the derivative of a constant is zero, so f'(x)=21x^2/5.

Answer 16

So, wait how are denominators and numerators switched?

Answer 17

Sorry. Took a food break.
Denominators and numerators? What got switched?

Answer 18

So I had it as f'(x)=-1/2+(21/5)x^2

Then you put it as f'(x)=21x^2/5.

Answer 19

yep, 21-over-5. Same as you. I just wrote it slightly differently.

Answer 20

\[\frac{21}{5}x^2 = \frac{21x^2}{5}\]

Answer 21

Oh. Okay. I have another question over finding the derivative of this problem:
\[f(x)=\frac{x ^{3}-3x^{2}+4}{ x^2 }\]

I have no clue how to do this one.

Answer 22

i. Divide out x^2 so you get f(x) = x - 3 + 4/(x^2)
Rewriting the last term, f(x) = \[x - 3 + 4x ^{-2}\]

Finding the derivative should be more manageable now