M mymath Oct 2012 5 0 Canada Oct 28, 2012 #1 Here is the problem: Use implicit differentiation to show (prove) that the derivative of y(x)=log _{22} (x) is y'(x) = 22^{x} 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.

Here is the problem: Use implicit differentiation to show (prove) that the derivative of y(x)=log _{22} (x) is y'(x) = 22^{x} 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.

C chiro MHF Helper Sep 2012 6,608 1,263 Australia Oct 28, 2012 #2 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). Reactions: 1 person

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).

S Soroban MHF Hall of Honor May 2006 12,028 6,341 Lexington, MA (USA) Oct 29, 2012 #3 Hello, mymath! I agree with Chiro. I suspect that the problem reads like this: \(\displaystyle \text{Using implicit differentiation,}\) \(\displaystyle \text{prove that the derivative of }\,{\color{red}y \:=\:22^x}\,\text{ is }\,y' \:=\: 22^x\ln(22)\) Click to expand... 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: .\(\displaystyle y' \:=\:y\cdot\ln(22) \quad\Rightarrow\quad y' \:=\:22^x\ln(22)\) Reactions: 1 person

Hello, mymath! I agree with Chiro. I suspect that the problem reads like this: \(\displaystyle \text{Using implicit differentiation,}\) \(\displaystyle \text{prove that the derivative of }\,{\color{red}y \:=\:22^x}\,\text{ is }\,y' \:=\: 22^x\ln(22)\) Click to expand... 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: .\(\displaystyle y' \:=\:y\cdot\ln(22) \quad\Rightarrow\quad y' \:=\:22^x\ln(22)\)

MarkFL Dec 2011 2,314 916 St. Augustine, FL. Oct 29, 2012 #6 A similar approach would be to write: \(\displaystyle y=22^x=e^{\ln(22^x)}=e^{x\ln(22)}\) Now, differentiate: \(\displaystyle \frac{dy}{dx}=e^{x\ln(22)}\ln(22)=\ln(22)22^x\)

A similar approach would be to write: \(\displaystyle y=22^x=e^{\ln(22^x)}=e^{x\ln(22)}\) Now, differentiate: \(\displaystyle \frac{dy}{dx}=e^{x\ln(22)}\ln(22)=\ln(22)22^x\)