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## AP®︎ Calculus AB (2017 edition)

### Course: AP®︎ Calculus AB (2017 edition)>Unit 4

Lesson 7: Derivatives of logarithmic functions

# Worked example: Derivative of log₄(x²+x) using the chain rule

Let's explore a worked example of differentiating the logarithmic function log₄(x²+x) using the chain rule. Leveraging our understanding of the derivative of logₐ(x), we simplify the complexities of composite functions, making differentiation more approachable and fun!

## Want to join the conversation?

• Hi, I don't quite understand how log base 4 of x is equal to 1/(ln4 * x). Could someone please explain this in more detail than he did in the video?
• In the video, he is talking about the differentiation of a logarithm with a different base. It is done like this:

You have:
y=log(base 4)(x)

Apply the change of base rule for logarithms:
y= log(base e)(x) / log(base e)(4)

I'll re-write so you can see it clearer (rewriting log(base e) as ln):
y=ln(x)/ln(4)
y= 1/ln(4) * ln(x)

Differentiate now:
dy/dx = 1/ln(4) * 1/x (since 1/ln(4) is just a constant)
dy/dx = 1/(xln(4) )
• I'm super confused here

why wouldn't the answer be ln((x^2)+x)/ln(4)?
(I'm using the "log base a times b=log base c times b / log base c times a" that he showed us earlier)

or if not that then 2x+1/ln(4)x
(taking derivatives of both and multiplying them together)

Could someone tell me which rules to use and when to use them?
thx so much!
• log_4 (x^2 + x) d/dx = ln(x^2+x)/ln(4) d/dx

let u = x^2 + x

=> y = ln(u)/ln(4)

=> dy/du = 1/(u*ln(4))

du/dx = 2x + 1

dy/du * du/dx = dy/dx

=> 1/(u*ln(4)) * (2x + 1) = (2x + 1)/(u*ln(4))

Substitute u = x^2 + x gives

(2x + 1) / (u * ln(4) ) = (2x+1) / ((x^2+x) * ln(4))

So you first you apply the log law that gives ln((x^2)+x)/ln(4). Then you just apply chain rule.

When apply chain rule on ln(f(x)) to differentiate that becomes

f'(x)/ f(x). So ln(f(x)) d/dx = f'(x)/f(x).
• Can someone please link the video Sal mentions about 'scaling the whole expression by lnx' ?
• Hi, I attempted to differentiate the log function by splitting composite function into two logs equations and differentiating them and then added the final two differentiated log equations and got 3/xln4. Why did this technique not work?
• It's not clear the steps that you took in this differentiation. Please be more specific.
• how is it composite function?
• The outer function is a logarithmic function and the inner function is a quadratic function. After all, log(x) is very different from log(ax^2+bx+c).
• What is the derivative of y = log x / (log (x) - 5) ?
• You can use the quotient rule. y'=(d/dx[log(x)] * (log(x) - 5) - d/dx[log(x) - 5] * log(x)) / (log(x)-5)^2 = -5/(ln(10)x(log(x) - 5)^2)
• What if the log base is smaller than e? Since that could make the derivative minus, even though the function may not be.
• That's no trouble. log₂(x²+x)=ln(x²+x)/ln(2), which is still positive. log(a) is negative only if a<1, and logarithms with bases less than 1 are undefined.
• @ shouldn't there be an x after ln4
(1 vote)
• for v(x) it would, but this is v(u(x)). The x in v(x) is replaced with u(x)
• What about functions like logbase4(x^2+1)^3 or logbase2(2x-4)^1/3? How would we find the derivative of them?
• (log_4 (x^2+1)^3 ) d/dx =

(log_e (x^2+1)^3 / log_e 4) d/dx

Since log_e 4 is just constant you can just factor it out.

To find the derivative of log_e (x^2+1)^3 use chain rule. You will often find many cases like expoential, trigonmetric, logarithmic, inverse trigonometric expressions in which you need to use chain rule so can find the derivative so you need to be comfortable with it.

Next substitute u= (x^2 + 1)^3, meaning du/dx = 6x(x^2 + 1)^3. Again you can you can use chain rule for the derivative of (x^2 + 1)^3 by substituting another variable or just taking the derivative of the inside and outside. And similarly since log_e u = y => dy/du = 1/u.

dy/du * du/dx = 1/u *6x(x^2 + 1)^3 = dy/dx. Obviously you need by log_e 4 for the actual derivative and the du cancels out in the multiplication as you`re dividing du by itself lim_u->0 u / lim_u->0 u where du = lim_u->0 u.