Implicit Differentiation

Oct 2012
5
0
Canada
Here is the problem:
Use implicit differentiation to show (prove) that the derivative of y(x)=log 22 (x) is y'(x) = 22x ln (22).

Here is what I have done so far:

y(x) = (lnx)/(ln 22)

Differentiating,
y'(x) = 1/(x ln 22).

but I do not get what is asked for.​
 

chiro

MHF Helper
Sep 2012
6,608
1,263
Australia
Hey mymath.

Your y'(x) is definitely not going to be that: I think the question is asking you to find the derivative of y(x) = 22^(x) instead of y(x) = log_22(x).
 
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Soroban

MHF Hall of Honor
May 2006
12,028
6,341
Lexington, MA (USA)
Hello, mymath!

I agree with Chiro.
I suspect that the problem reads like this:


\(\displaystyle \text{Using implicit differentiation,}\)
\(\displaystyle \text{prove that the derivative of }\,{\color{red}y \:=\:22^x}\,\text{ is }\,y' \:=\: 22^x\ln(22)\)

We have: .\(\displaystyle y \;=\:22^x\)

Take logs: .\(\displaystyle \ln(y) \:=\:\ln(22^x) \quad\Rightarrow\quad \ln(y) \:=\:x\cdot\ln(22)\)

Differentiate: .\(\displaystyle \frac{1}{y}\!\cdot\!y' \:=\:1\cdot\ln(22)\)

Therefore: .\(\displaystyle y' \:=\:y\cdot\ln(22) \quad\Rightarrow\quad y' \:=\:22^x\ln(22)\)

 
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Oct 2012
5
0
Canada
yes, Soroban, you are right.

y(x) = 22^x.

Thanks a lot.
 
Oct 2012
5
0
Canada
Chiro, you are right.

y(x) = 22^x.

Thank you.
 
Dec 2011
2,314
916
St. Augustine, FL.
A similar approach would be to write:

\(\displaystyle y=22^x=e^{\ln(22^x)}=e^{x\ln(22)}\)

Now, differentiate:

\(\displaystyle \frac{dy}{dx}=e^{x\ln(22)}\ln(22)=\ln(22)22^x\)