Math / Calculus

Question1 point)
Differentiate $f(x)=\log _{2}(4-5 x)$ $f^{\prime}(x)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a logarithmic function with base 2. Watch out! Remember the change of base formula and the chain rule!

STEP 2

1. Rewrite the logarithm
2. Differentiate the function

STEP 3

We'll rewrite our function $f(x) = \log_2(4 - 5x)$ using the change of base formula.
This makes it easier to differentiate because we're more used to working with natural logarithms!
Remember, the change of base formula says $\log_a(b) = \frac{\log_c(b)}{\log_c(a)}$ .
Here, we'll change to base e (natural logarithm): $f(x) = \frac{\ln(4 - 5x)}{\ln(2)}$ Remember $\ln(x)$ just means $\log_e(x)$ .

STEP 4

Notice that $\ln(2)$ is just a constant.
This means we can rewrite our function as: $f(x) = \frac{1}{\ln(2)} \cdot \ln(4 - 5x)$ This makes it super clear how to differentiate!

STEP 5

Now, we'll differentiate $f(x)$ using the chain rule.
Remember, the chain rule says that the derivative of a composite function is the derivative of the outer function (evaluated at the inner function) times the derivative of the inner function.
Here, our outer function is $\ln(u)$ (where $u = 4 - 5x$ ) and the derivative is $\frac{1}{u}$ .
Our inner function is $4 - 5x$ and its derivative is $-5$ .
So, we have: $f'(x) = \frac{1}{\ln(2)} \cdot \frac{1}{4 - 5x} \cdot (-5)$

STEP 6

Let's clean things up a bit: $f'(x) = \frac{-5}{\ln(2) \cdot (4 - 5x)}$ Look at that, nice and tidy!

STEP 7

The derivative of $f(x) = \log_2(4 - 5x)$ is $f'(x) = \frac{-5}{\ln(2)(4 - 5x)}$ .

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Question1 point)
Differentiate $f(x)=\log _{2}(4-5 x)$ $f^{\prime}(x)=$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a logarithmic function with base 2. Watch out! Remember the change of base formula and the chain rule!

STEP 2

1. Rewrite the logarithm
2. Differentiate the function

STEP 3

STEP 4

Notice that $\ln(2)$ is just a constant.
This means we can rewrite our function as: $f(x) = \frac{1}{\ln(2)} \cdot \ln(4 - 5x)$ This makes it super clear how to differentiate!

STEP 5

STEP 6

Let's clean things up a bit: $f'(x) = \frac{-5}{\ln(2) \cdot (4 - 5x)}$ Look at that, nice and tidy!

STEP 7

The derivative of $f(x) = \log_2(4 - 5x)$ is $f'(x) = \frac{-5}{\ln(2)(4 - 5x)}$ .

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Question1 point)Differentiate f(x)=log⁡2(4−5x)f(x)=\log _{2}(4-5 x)f(x)=log2​(4−5x) f′(x)=f^{\prime}(x)=f′(x)= □\square□

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a logarithmic function with base 2. Watch out! Remember the change of base formula and the chain rule!

STEP 2

1. Rewrite the logarithm2. Differentiate the function

STEP 3

STEP 4

Notice that ln⁡(2)\ln(2)ln(2) is just a **constant**.This means we can rewrite our function as: f(x)=1ln⁡(2)⋅ln⁡(4−5x)f(x) = \frac{1}{\ln(2)} \cdot \ln(4 - 5x)f(x)=ln(2)1​⋅ln(4−5x) This makes it super clear how to **differentiate**!

STEP 5

STEP 6

Let's **clean things up** a bit: f′(x)=−5ln⁡(2)⋅(4−5x)f'(x) = \frac{-5}{\ln(2) \cdot (4 - 5x)}f′(x)=ln(2)⋅(4−5x)−5​ Look at that, nice and tidy!

STEP 7

The derivative of f(x)=log⁡2(4−5x)f(x) = \log_2(4 - 5x)f(x)=log2​(4−5x) is f′(x)=−5ln⁡(2)(4−5x)f'(x) = \frac{-5}{\ln(2)(4 - 5x)}f′(x)=ln(2)(4−5x)−5​.

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Question1 point)
Differentiate $f(x)=\log _{2}(4-5 x)$ $f^{\prime}(x)=$ $\square$

1. Rewrite the logarithm
2. Differentiate the function

Notice that $\ln(2)$ is just a constant.
This means we can rewrite our function as: $f(x) = \frac{1}{\ln(2)} \cdot \ln(4 - 5x)$ This makes it super clear how to differentiate!

Let's clean things up a bit: $f'(x) = \frac{-5}{\ln(2) \cdot (4 - 5x)}$ Look at that, nice and tidy!

The derivative of $f(x) = \log_2(4 - 5x)$ is $f'(x) = \frac{-5}{\ln(2)(4 - 5x)}$ .