# Thread: Derivatives and Tangent lines

1. ## Derivatives and Tangent lines

As always, I have a very loose definition of what these things mean, so going slow would be appreciated.

1) g(x) = sqrt(9 + 4x)

2) 9cos(x) + (5/7)cot(x)

3) y = (6 + sin(x)) / (6x + cos(x)) Find the derivative

4) sec(x) Find the second derivative fprime(pi/3)

5) y = e^(9e^x))

6) y = sqrt( (1-x) / (1 + x))

I really don't have a clue what I'm doing here.

2. Originally Posted by CFem
As always, I have a very loose definition of what these things mean, so going slow would be appreciated.

1) g(x) = sqrt(9 + 4x)

chain rule

2) 9cos(x) + (5/7)cot(x)

straight-forward derivative of two trig functions

3) y = (6 + sin(x)) / (6x + cos(x)) Find the derivative

quotient rule

4) sec(x) Find the second derivative fprime(pi/3)

derivative of the derivative ... 2nd derivative will involve the product rule.

5) y = e^(9e^x))

chain rule again

6) y = sqrt( (1-x) / (1 + x))

chain rule in conjunction with the quotient rule ... unless you know logarithmic differentiation.

I really don't have a clue what I'm doing here.

then get one ... go to the link.
Pauls Online Notes : Calculus I - Derivatives

3. I think I've messed up the Chain Rule somehow.

Working with number 1, I ended up with 1/(18 + 8x)^(1/2)

I cut the function up into sqrt(x) and 9 + 4x
Derivatives being (1/2)x^(1/2) and 4 respectively

=> ((1/2)( 9 + 4x)^(1/2)) * 4

Where did I go wrong?

4. Originally Posted by CFem
I think I've messed up the Chain Rule somehow.

Working with number 1, I ended up with 1/(18 + 8x)^(1/2)

I cut the function up into sqrt(x) and 9 + 4x
Derivatives being (1/2)x^(1/2) and 4 respectively

=> ((1/2)( 9 + 4x)^(1/2)) * 4

Where did I go wrong?
you cannot "split" two terms inside a radical ... $\displaystyle \sqrt{a+b} \ne \sqrt{a} + \sqrt{b}$ ... review your algebra.

$\displaystyle y = \sqrt{9+4x}$

$\displaystyle y = (9+4x)^{\frac{1}{2}}$

$\displaystyle \frac{dy}{dx} = \frac{1}{2}(9+4x)^{-\frac{1}{2}} \cdot (4)$

$\displaystyle \frac{dy}{dx} = 2(9+4x)^{-\frac{1}{2}} = \frac{2}{\sqrt{9+4x}}$

5. That's what I had, I just don't think it translated well over text.

But thanks. I lost a negative sign it looks like.

6. Sorry if double posting isn't allowed, but I didn't think it would be necessary to make another topic.

On number 3:

f(x) = 6 + sin(x)
f'(x) = cos(x)

g(x) = 6x + cos(x)
g'(x) = 6 - sin(x)

(6x + cos(x))(cos(x)) - (6 + sin(x))(6 - sin(x))
-------------------------------------
(6 + cos(x))^2

6xcos(x) + cos(x)^2 -36 + sin(x)^2
---------------------------------
36 + 12cos(x) + cos(x)^2

=

6xcosx + sin(x)^2
-----------------
12cos(x)

I have a feeling I made another simple mistake or I forgot to simplify something...

7. Originally Posted by CFem
Sorry if double posting isn't allowed, but I didn't think it would be necessary to make another topic.

On number 3:

f(x) = 6 + sin(x)
f'(x) = cos(x)

g(x) = 6x + cos(x)
g'(x) = 6 - sin(x)

(6x + cos(x))(cos(x)) - (6 + sin(x))(6 - sin(x))
-------------------------------------
(6 + cos(x))^2

6xcos(x) + cos(x)^2 -36 + sin(x)^2
---------------------------------
36 + 12cos(x) + cos(x)^2

... you cannot "cancel" the cos^2(x)'s , that is an algebra no-no.
... note (in the numerator) that cos^2(x) + sin^2(x) = 1
...

8. Clearly I am not understanding this

9. Originally Posted by CFem
Clearly I am not understanding this
your calculus on the last question was fine ... it's your algebra that is hurting you.

10. 6xcos(x) + 1 - 1
----------------
12cos(x) + cos(x)^2

It comes back to Trig more than anything.

Not sure where to go from here.

11. Originally Posted by CFem
On number 3:

f(x) = 6 + sin(x)
f'(x) = cos(x)

g(x) = 6x + cos(x)
g'(x) = 6 - sin(x)

(6x + cos(x))(cos(x)) - (6 + sin(x))(6 - sin(x))
-------------------------------------
(6x + cos(x))^2 ... leave this alone, squaring it does little to help.

6xcos(x) + cos(x)^2 - 36 + sin(x)^2
---------------------------------
[6x + cos(x)]^2

6xcosx + 1 - 36
-----------------
[6x + cos(x)]^2

6xcosx - 35
---------------
[6x + cos(x)]^2
you're done.

12. Our homework gets checked by some sort of automated system, which spat that out.

Thanks anyway.

13. 3) F(x) = e^(9e^x))

f(x) = e^x
f'(x) = e^x

g(x) = 9e^x
g'(x) = 9(d/dx)e^x = 9e^x

F'(x) = f'(g(x))(g'(x))

=> e^(9^e^x) * 9e^x
=> 9e^(9xe^x)

More Algebra problems?

something) Find the tangent line of x^4 + 8e^x

I ended up using L'Hospital's Rule and got the slope of the tangent line to = 8 (0 without the rule) and ended up with y = -8(x-1), but I lost my work on this one. Ideas?

14. the general derivative for $\displaystyle y = e^u$, where $\displaystyle u$ is a function of $\displaystyle x$ ...

$\displaystyle \frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}$

for $\displaystyle f(x) = e^{9e^x}$ , the $\displaystyle u$ is $\displaystyle 9e^x$ ... $\displaystyle \frac{du}{dx} = 9e^x$

so ... $\displaystyle f'(x) = e^{9e^x} \cdot 9e^x = 9e^{9e^x} \cdot e^x = 9e^{9e^x+x}$

15. Originally Posted by CFem
something) Find the tangent line of x^4 + 8e^x

I ended up using L'Hospital's Rule and got the slope of the tangent line to = 8 (0 without the rule) and ended up with y = -8(x-1), but I lost my work on this one. Ideas?
tangent line at what point?

L'Hopital is for limits ... ???

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