# Differentiation

• Dec 30th 2007, 08:20 AM
DRJ
Differentiation
I am currently studying calculus on an open learning basis and was wondering if someone could take a look at these questions and answers I have posted and feed back on my method of working out etc. I am pretty sure these answers are correct but would appreciate any comments.

Question 1:

Obtain $\frac{dy}{dx}$ for the following expression.

$y=(5x + 4)^3$ so $3(5x + 4)^2$ This is the 1st part.

I then differentiate the inner part of the function so:-

$(5x + 4)$ differentiates to 5. This is the the 2nd part.

I then multiply the 1st by the 2nd to get $3(5x + 4)^2\times 5$

so $\frac{dy}{dx}=15(5x + 4)^2$

Question 2:

Obtain $\frac{dy}{dx}$ for the following expression.

$y=(3 - 2x)^5$ so $5(3 - 2x)^4$ This is the 1st part.

I then differentiate the inner part of the function so:-

$(3 - 2x)$ differentiates to -2. This is the the 2nd part.

I then multiply the 1st by the 2nd to get $5(3 - 2x)^4\times -2$

so $\frac{dy}{dx}=-10(3 - 2x)^4$

Question 3:

Obtain $\frac{dy}{dx}$ for the following expression.

$y=\sqrt{(5 - 0.6x)}$ which is $(5 - 0.6x)^\frac{1}{2}$

I then differentiate this to get $\frac{1}{2}(5 - 0.6x)^{-\frac{1}{2}}$ This is the 1st part:

$(5 - 0.6x)$ differentiates to -0.6. This is the the 2nd part.

I then multiply the 1st by the 2nd to get $\frac{1}{2}(5 - 0.6x)^{-\frac{1}{2}}\times -0.6$

so $\frac{dy}{dx}=-0.3(5 - 0.6x)^{-\frac{1}{2}}$

Question 4:

Obtain $\frac{dy}{dx}$ for the following expression.

$y=(2 + 3x)^-0.6$ so $-0.6(2 + 3x)^{-1.6}$ This is the 1st part.

I then differentiate the inner part of the function so:-

$(2 + 3x)$ differentiates to 3. This is the the 2nd part.

I then multiply the 1st by the 2nd to get $-0.6(2 + 3x)^{-1.6}\times 3$

so $\frac{dy}{dx}=-1.8(2 + 3x)^{-1.6}$

As I said at the beginning I am pretty sure I am 100% correct and I agree these are relatively simple questions from a calculus point of view but being new to this subject I was just wondering if someone could make comment please.

Thanks

Derek.
• Dec 30th 2007, 08:26 AM
Isomorphism
I think they are all right :D

And there is nothing much to comment either... Why not try harder ones??
• Dec 31st 2007, 07:09 AM
DRJ
Obtain the derivative of
I came across another area which I am trying to get to grips with and was wondering if someone could help

I need to obtain the derivative of $f(t) = ln(5 - \frac{2}{3}t)$

I think I procede as follows.

$\frac{dy}{dt} = \frac{1}{5 - \frac{2}{3}t}\times -\frac{2}{3}$

Finishing up with $f'(t) = \frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$

or $\frac{-0.6}{5 - 0.6 t}$

Another one is Determine $f'(y)$if $f(y) = exp(3 - \frac{1}{4}y)$

When the question reads "exp" do I assumeit is refering to the exponential function $e$

I have hunted high and low for similar equations to this to give me some kind of example to work too but come up with more or less nothing so from what I have found I have tried with this. I am probably wrong. Could someone make comment please??

Thanks

Derek
• Dec 31st 2007, 07:45 AM
Isomorphism
Quote:

Originally Posted by DRJ
Another one is Determine $f'(y)$if $f(y) = exp(3 - \frac{1}{4}y)$

When the question reads "exp" do I assumeit is refering to the exponential function $e$

I have hunted high and low for similar equations to this to give me some kind of example to work too but come up with more or less nothing so from what I have found I have tried with this. I am probably wrong. Could someone make comment please??

Thanks

Derek

If $f(x) = e^{ax+b}$ then $f'(x) = ae^{ax+b}$
• Dec 31st 2007, 07:46 AM
topsquark
Quote:

Originally Posted by DRJ
I came across another area which I am trying to get to grips with and was wondering if someone could help

I need to obtain the derivative of $f(t) = ln(5 - \frac{2}{3}t)$

I think I procede as follows.

$\frac{dy}{dt} = \frac{1}{5 - \frac{2}{3}t}\times -\frac{2}{3}$

Finishing up with $f'(t) = \frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$

or $\frac{-0.6}{5 - 0.6 t}$

Yes, though if this is a Math class I'd recommend leaving this as
$f^{\prime}(t) = -\frac{2}{15 - 2t}$
rather than the decimal form.

Quote:

Originally Posted by DRJ
When the question reads "exp" do I assumeit is refering to the exponential function $e$

Yes, the "exp" and "e" exponential are the same function.

-Dan
• Dec 31st 2007, 09:34 AM
Soroban
Hello, DRJ!

Quote:

Find the derivative of: $f(t) \:= \:\ln\left(5 - \frac{2}{3}t\right)$

I think I proceed as follows: . $\frac{dy}{dt} \:= \:\frac{1}{5 - \frac{2}{3}t} \times \left(-\frac{2}{3}\right)$

Finishing up with: . $f'(t) \:= \:\frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$ . . . . Yes!

. . or: . $\frac{-0.6}{5 - 0.6 t}$ . . . . no

Don't use rounded-off decimals . . . $0.6$ is not equal to $\frac{2}{3}$

We have: . $f'(x) \:=\:\frac{-\frac{2}{3}}{5 - \frac{2}{3}t}$

Multiply top and bottom by 3: . $\frac{3 \times\left(-\frac{2}{3}\right)}{3 \times\left(5 - \frac{2}{3}t\right)} \;=\;\frac{-2}{15-2t}$

Quote:

Determine $f'(y)\text{ if }f(y) \:= \:\text{exp}\left(3 - \frac{1}{4}y\right)$

do I assume it is refering to the exponential function $e$ ? . Yes!

. . . . . . . . . . . . . . the original function
. . . . . . . . . . . . . . . . . . . $\downarrow$
Given: . $y \:=\:e^u$, then: . $y' \:=\:e^u\cdot u'$
. - . - . - . . . . . . . . . . . . . . . $\uparrow$
. . . . . . . . . . . . . . .
derivative of the exponent

We have: . $f(y) \:=\;e^{(3-\frac{1}{4}y)}$

Then: . $f'(y) \;=\;e^{(3-\frac{1}{4}y)}\!\cdot\!\left(\text{-}\frac{1}{4}\right) \;=\;-\frac{1}{4}e^{(3-\frac{1}{4}y)}$

• Jan 1st 2008, 03:14 AM
DRJ
Differentiation and the exponential function
Hello again

Just a quick query regarding the $e$ function.

In my example $f(y) = exp(3 - \frac{1}{4}y)$ I think I may have mis read the question.

Is this actually asking me to find $f(y)$ of $exp^{(3-\frac{1}{4}y)}$ or as you stated $e^{(3-\frac{1}{4}y)}$ as opposed to $f(y) = exp(3 - \frac{1}{4}y)$ or are they both the same???

Thanks and happy new year etc
• Jan 1st 2008, 04:35 AM
topsquark
Quote:

Originally Posted by DRJ
Hello again

Just a quick query regarding the $e$ function.

In my example $f(y) = exp(3 - \frac{1}{4}y)$ I think I may have mis read the question.

Is this actually asking me to find $f(y)$ of $exp^{(3-\frac{1}{4}y)}$ or as you stated $e^{(3-\frac{1}{4}y)}$ as opposed to $f(y) = exp(3 - \frac{1}{4}y)$ or are they both the same???

Thanks and happy new year etc

There is no such thing as $exp^x$. exp(x) is the operation $e^x$, so there is no room for confusion.

-Dan