Differential cAlculus - Implicit Differentiation

[math1]

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.

 

[tkhunny]

There is nothing "implicit" about what you have written unless you have misunderstood what you have written.

I'm officially guessing that y is some function of x and the expression y(x) is suppose to mean y*x, Thus

Try this problem: y*x = log₂₂(x)

 

[math1]

no,
it is same as I have written as

y(x) = log₂₂(x).

 

[daon2]

The question is asking you to do something which is not possible.

Had the question been "y(x)=22^x, show y'(x) = 22^x*ln(22)" then you'd be golden

 

[HallsofIvy]

no,
it is same as I have written as

y(x) = log₂₂(x).

So that \(\displaystyle 22^y= x\).

Now, differentiate both sides of that with respect to x.

 

[math1]

So that \(\displaystyle 22^y= x\).

Now, differentiate both sides of that with respect to x.

By differentiating, I got,

22^y. ln 22 .y' = 1

y' = 1/(22^y. ln 22)

y' = 1/(x. ln 22)

which is not as required.

 

[mmm4444bot]

Daon2 already explained that the given information is wrong. Please speak with your instructor.

log₂₂(x) is not an anti-derivative of 22^x*ln(22)

The derivative of log₂₂​(x) is 1/(x*ln(22))


Note re notation: The symbols x and X do not represent the same number. Do not interchange x and X.