# Implicit Differentiation

#### mymath

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
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).

• 1 person

#### Soroban

MHF Hall of Honor
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: .$$\displaystyle y' \:=\:y\cdot\ln(22) \quad\Rightarrow\quad y' \:=\:22^x\ln(22)$$

• 1 person

#### mymath

yes, Soroban, you are right.

y(x) = 22^x.

Thanks a lot.

#### mymath

Chiro, you are right.

y(x) = 22^x.

Thank you.

#### MarkFL

A similar approach would be to write:

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

Now, differentiate:

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