# Differentiate?

• March 28th 2010, 11:27 PM
MathKidd
1.Equation: $f(x)=3ln(3-7x)$
Solution: [tex]f'(x)=(3*-7)*\frac{u'}{u}
= $f'(x)=\frac{-21}{2-7x}$

2.Equation: $f(x)=e^{5x+2}$
solution: $f'(x)=5e^{5x+2}$

3. Equation: $g(x)=7^{x-3}+25$
Solution: $g'(x)=7^{x-3}Log7$

4. Equation: $f(x)=Log3x-Ln3$
Solution: $f'(x)=\frac{1}{x Log3}$

5. Equation: $h(x)=log2(5x+4)$
Solution: $h'(x)=\frac{5}{(5x+4)(log2)}$

6. Equation: $y= Ln \frac{(x-2)^4}{(1+x)^5}$
Solution: Unsure of how to solve

Derivative of

1. Equation= $a^x$
Solution=???

2 Equation= $e^u$
Solution= $a^u*u^1 = u^1*e^x$ or is it $e^u$

3 Equation = $a^u$
solution = ???

4. Find Result
Equation $Ln*\frac{u^4*}{z^6}$
Solution = $\frac{1}{x}*\frac{(u^4*v)(6z^5)-(4u^5)(z^6)}{(z^6)^2}$
• March 29th 2010, 12:16 AM
harish21
Quote:

Originally Posted by MathKidd
Is differentiate the same as derivatives? and is this question correct?

Problem: $f(x)=e$ Raised to the power of or ^ (5x+2)
Solution: = $5e$ Raised to the power of or ^ (5x+2)
or would it be written as $f1(x)=5e$ Raised to the power of or ^ (5x+2) (Assuming differentiate and derivatives are the same.)

Yes differentiate is the same as finding the derivative, Now to your question:

$f(x)=e^{5x+2}$. Your latex input seems incorrect, but I guess this is what you are asking.

so,

$f'(x) = 5 \times e^{5x+2}$

• March 29th 2010, 01:15 AM
MathKidd
Ok...
• March 29th 2010, 04:35 AM
HallsofIvy
$a^x= e^{ln a^x}= e^{x ln a}$.

Therefore, $(a^x)'= e^{x ln a}(ln x)= a^x ln a$.

Notice that $e^x$ is "special" only because ln e= 1.
• March 29th 2010, 07:55 AM
MathKidd
are my other solutions valid?
• March 29th 2010, 08:08 AM
ione
Quote:

Originally Posted by MathKidd
6. Equation: $y= Ln \frac{(x-2)^4}{(1+x)^5}$
Solution: Unsure of how to solve

$y= ln \frac{(x-2)^4}{(1+x)^5}$

$=4ln(x-2)-5ln(1+x)$

Now can you find the derivative?

Quote:

Originally Posted by MathKidd
4. Find Result
Equation $Ln\frac{u^4*}{z^6}$
Solution $= \frac{1}{x}*\frac{(u^4*v)(6z^5)-(4u^5)(z^6)}{(z^6)^2}$

This is like #6 above

Quote:

Originally Posted by MathKidd
5. Equation: $h(x)=log2(5x+4)$
Solution: $h'(x)=\frac{5}{(5x+4)(log2)}$

$h(x)=log_2(5x+4)$

$h'(x)=\frac{5}{(5x+4)(ln2)}$