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February 20th, 2016, 11:55 PM   #1
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Derivatives of Exponential and Log Functions

Hi Guys,

I just solved the questions below and need some feedback. Just want to make sure, I solved them correctly. Thank you so much.


1. $\displaystyle y = 0.5e^x-3.17e^2x$

Solution: $\displaystyle y' = 0.5e^x(1) - 2(3.17)e^2x$
$\displaystyle y' = 0.5e^x-6.34e^2x$
====================================

2. $\displaystyle 2e^{5x^2-4x}$
Solution: $\displaystyle g'(x) = 10x-4 $
$\displaystyle y' = 2(10x-4)e^{5x^2-4x} $
==========================================

3. [MATH]y = ln(5x^4-x^3)^2[/MATH
Solution: $\displaystyle y' = 2(5x^4-x^3).(20x^3-3x^2)/(5x^4-x^3)^2$
===========================================

4. $\displaystyle y=ln\frac {7x-1}{12x+5}$
Solution: $\displaystyle g'(x) = \frac{7(12x+5)-(7x-1)(12)}{(12x+5)^2}$
$\displaystyle = \frac{84x+35-84x+12}{(12x+5)^2}$
$\displaystyle = \frac{47}{(12x+5)^2}$
$\displaystyle \frac{g'(x)}{g(x)} = \frac{\frac{47}{(12x+5)^2}}{ \frac{7x-1}{12x+5}}$
$\displaystyle = \frac{47(12x+5)}{(12x+5)^2(7x-1)}$
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February 21st, 2016, 12:10 AM   #2
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1. There's no exponential function in the second term. $\displaystyle 3.17e^2$ is the coefficient of x.

2. I think you're correct, but you may want to factorise the expression in brackets (if your teacher requires it).

3. I think you missed a ln in the first part. In the second part, you should cancel out common factors.

4. Again, you should cancel out the common factors.
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February 21st, 2016, 12:25 AM   #3
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Quote:
Originally Posted by 123qwerty View Post
1. There's no exponential function in the second term. $\displaystyle 3.17e^2$ is the coefficient of x.

2. I think you're correct, but you may want to factorise the expression in brackets (if your teacher requires it).

3. I think you missed a ln in the first part. In the second part, you should cancel out common factors.

4. Again, you should cancel out the common factors.
Actually for no. 1 its not, I typed it wrong, it should have been:

$\displaystyle y=0.5e^x−3.17e^{2x}$
$\displaystyle y′=0.5e^x(1)−2(3.17)e^{2x}$
$\displaystyle y′=0.5e^x−6.34e^{2x}$

Is that correct now?

By the way, greatly appreciate your quick response. Thank You.
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February 21st, 2016, 12:39 AM   #4
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Quote:
Originally Posted by atari View Post
Actually for no. 1 its not, I typed it wrong, it should have been:

$\displaystyle y=0.5e^x−3.17e^{2x}$
$\displaystyle y′=0.5e^x(1)−2(3.17)e^{2x}$
$\displaystyle y′=0.5e^x−6.34e^{2x}$

Is that correct now?

By the way, greatly appreciate your quick response. Thank You.
Yep
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