# Differentiate?

• Mar 28th 2010, 10:27 PM
MathKidd
1.Equation: $\displaystyle f(x)=3ln(3-7x)$
Solution: [tex]f'(x)=(3*-7)*\frac{u'}{u}
=$\displaystyle f'(x)=\frac{-21}{2-7x}$

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

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

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

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

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

Derivative of

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

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

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

4. Find Result
Equation $\displaystyle Ln*\frac{u^4*}{z^6}$
Solution = $\displaystyle \frac{1}{x}*\frac{(u^4*v)(6z^5)-(4u^5)(z^6)}{(z^6)^2}$
• Mar 28th 2010, 11:16 PM
harish21
Quote:

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

Problem:$\displaystyle f(x)=e$ Raised to the power of or ^ (5x+2)
Solution: = $\displaystyle 5e$ Raised to the power of or ^ (5x+2)
or would it be written as $\displaystyle f1(x)=5e$ 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:

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

so,

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

• Mar 29th 2010, 12:15 AM
MathKidd
Ok...
• Mar 29th 2010, 03:35 AM
HallsofIvy
$\displaystyle a^x= e^{ln a^x}= e^{x ln a}$.

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

Notice that $\displaystyle e^x$ is "special" only because ln e= 1.
• Mar 29th 2010, 06:55 AM
MathKidd
are my other solutions valid?
• Mar 29th 2010, 07:08 AM
ione
Quote:

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

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

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

Now can you find the derivative?

Quote:

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

This is like #6 above

Quote:

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

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

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