Math / Calculus

QuestionDifferentiate the following function. $\begin{array}{l} y=(\ln x)^{8}+\ln \left(x^{8}\right) \\ \frac{d y}{d x}=\square \end{array}$ $\square$

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that has two parts: one is the natural logarithm of $x$ raised to the power of 8, and the other is the natural logarithm of $x$ raised to the power of 8. Watch out! Don't mix up the power rule and the chain rule!
Also, remember the properties of logarithms!

STEP 2

1. Rewrite the function
2. Differentiate the first term
3. Differentiate the second term
4. Combine the results

STEP 3

Let's rewrite our function $y$ to make it easier to differentiate.
We can use the power rule of logarithms: $\ln(a^b) = b \cdot \ln(a)$ .
So, $\ln(x^8)$ becomes $8 \cdot \ln(x)$ .

STEP 4

Our rewritten function now looks like this: $y = (\ln(x))^8 + 8 \cdot \ln(x)$ Much cleaner!

STEP 5

Now, let's differentiate $(\ln(x))^8$ .
We'll need the chain rule here.
Think of it like this: we have an inner function, $\ln(x)$ , and an outer function, something raised to the power of 8.

STEP 6

The chain rule says: derivative of the outer function (with the inner function left alone) times the derivative of the inner function.

STEP 7

The derivative of the outer function is $8(\ln(x))^7$ .
The derivative of the inner function, $\ln(x)$ , is $\frac{1}{x}$ .

STEP 8

Multiplying these together gives us: $8(\ln(x))^7 \cdot \frac{1}{x} = \frac{8(\ln(x))^7}{x}$

STEP 9

Next, let's differentiate $8 \cdot \ln(x)$ .
Since 8 is a constant, we can just multiply it by the derivative of $\ln(x)$ , which is $\frac{1}{x}$ .

STEP 10

This gives us: $8 \cdot \frac{1}{x} = \frac{8}{x}$

STEP 11

Finally, we add the derivatives of both terms to find the derivative of the whole function: $\frac{dy}{dx} = \frac{8(\ln(x))^7}{x} + \frac{8}{x}$ We can also write this as: $\frac{dy}{dx} = \frac{8(\ln(x))^7 + 8}{x}$

STEP 12

$\frac{dy}{dx} = \frac{8(\ln(x))^7 + 8}{x}$

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QuestionDifferentiate the following function. $\begin{array}{l} y=(\ln x)^{8}+\ln \left(x^{8}\right) \\ \frac{d y}{d x}=\square \end{array}$ $\square$

Studdy Solution

STEP 1

STEP 2

1. Rewrite the function
2. Differentiate the first term
3. Differentiate the second term
4. Combine the results

STEP 3

Let's rewrite our function $y$ to make it easier to differentiate.
We can use the power rule of logarithms: $\ln(a^b) = b \cdot \ln(a)$ .
So, $\ln(x^8)$ becomes $8 \cdot \ln(x)$ .

STEP 4

Our rewritten function now looks like this: $y = (\ln(x))^8 + 8 \cdot \ln(x)$ Much cleaner!

STEP 5

Now, let's differentiate $(\ln(x))^8$ .
We'll need the chain rule here.
Think of it like this: we have an inner function, $\ln(x)$ , and an outer function, something raised to the power of 8.

STEP 6

The chain rule says: derivative of the outer function (with the inner function left alone) times the derivative of the inner function.

STEP 7

The derivative of the outer function is $8(\ln(x))^7$ .
The derivative of the inner function, $\ln(x)$ , is $\frac{1}{x}$ .

STEP 8

Multiplying these together gives us: $8(\ln(x))^7 \cdot \frac{1}{x} = \frac{8(\ln(x))^7}{x}$

STEP 9

Next, let's differentiate $8 \cdot \ln(x)$ .
Since 8 is a constant, we can just multiply it by the derivative of $\ln(x)$ , which is $\frac{1}{x}$ .

STEP 10

This gives us: $8 \cdot \frac{1}{x} = \frac{8}{x}$

STEP 11

Finally, we add the derivatives of both terms to find the derivative of the whole function: $\frac{dy}{dx} = \frac{8(\ln(x))^7}{x} + \frac{8}{x}$ We can also write this as: $\frac{dy}{dx} = \frac{8(\ln(x))^7 + 8}{x}$

STEP 12

$\frac{dy}{dx} = \frac{8(\ln(x))^7 + 8}{x}$

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QuestionDifferentiate the following function. y=(ln⁡x)8+ln⁡(x8)dydx=□\begin{array}{l} y=(\ln x)^{8}+\ln \left(x^{8}\right) \\ \frac{d y}{d x}=\square \end{array}y=(lnx)8+ln(x8)dxdy​=□​ □\square□

Studdy Solution

STEP 1

STEP 2

1. Rewrite the function2. Differentiate the first term3. Differentiate the second term4. Combine the results

STEP 3

Let's **rewrite** our function yyy to make it easier to differentiate.We can use the **power rule of logarithms**: ln⁡(ab)=b⋅ln⁡(a)\ln(a^b) = b \cdot \ln(a)ln(ab)=b⋅ln(a).So, ln⁡(x8)\ln(x^8)ln(x8) becomes 8⋅ln⁡(x)8 \cdot \ln(x)8⋅ln(x).

STEP 4

Our **rewritten function** now looks like this: y=(ln⁡(x))8+8⋅ln⁡(x)y = (\ln(x))^8 + 8 \cdot \ln(x)y=(ln(x))8+8⋅ln(x) Much cleaner!

STEP 5

Now, let's **differentiate** (ln⁡(x))8(\ln(x))^8(ln(x))8.We'll need the **chain rule** here.Think of it like this: we have an **inner function**, ln⁡(x)\ln(x)ln(x), and an **outer function**, something raised to the power of 8.

STEP 6

The chain rule says: **derivative of the outer function (with the inner function left alone) times the derivative of the inner function**.

STEP 7

The **derivative of the outer function** is 8(ln⁡(x))78(\ln(x))^78(ln(x))7.The **derivative of the inner function**, ln⁡(x)\ln(x)ln(x), is 1x\frac{1}{x}x1​.

STEP 8

Multiplying these together gives us: 8(ln⁡(x))7⋅1x=8(ln⁡(x))7x8(\ln(x))^7 \cdot \frac{1}{x} = \frac{8(\ln(x))^7}{x}8(ln(x))7⋅x1​=x8(ln(x))7​

STEP 9

Next, let's **differentiate** 8⋅ln⁡(x)8 \cdot \ln(x)8⋅ln(x).Since 8 is a **constant**, we can just multiply it by the **derivative of** ln⁡(x)\ln(x)ln(x), which is 1x\frac{1}{x}x1​.

STEP 10

This gives us: 8⋅1x=8x8 \cdot \frac{1}{x} = \frac{8}{x}8⋅x1​=x8​

STEP 11

STEP 12

dydx=8(ln⁡(x))7+8x\frac{dy}{dx} = \frac{8(\ln(x))^7 + 8}{x}dxdy​=x8(ln(x))7+8​

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QuestionDifferentiate the following function. $\begin{array}{l} y=(\ln x)^{8}+\ln \left(x^{8}\right) \\ \frac{d y}{d x}=\square \end{array}$ $\square$

1. Rewrite the function
2. Differentiate the first term
3. Differentiate the second term
4. Combine the results

Let's rewrite our function $y$ to make it easier to differentiate.
We can use the power rule of logarithms: $\ln(a^b) = b \cdot \ln(a)$ .
So, $\ln(x^8)$ becomes $8 \cdot \ln(x)$ .

Our rewritten function now looks like this: $y = (\ln(x))^8 + 8 \cdot \ln(x)$ Much cleaner!

Now, let's differentiate $(\ln(x))^8$ .
We'll need the chain rule here.
Think of it like this: we have an inner function, $\ln(x)$ , and an outer function, something raised to the power of 8.

The chain rule says: derivative of the outer function (with the inner function left alone) times the derivative of the inner function.

The derivative of the outer function is $8(\ln(x))^7$ .
The derivative of the inner function, $\ln(x)$ , is $\frac{1}{x}$ .

Multiplying these together gives us: $8(\ln(x))^7 \cdot \frac{1}{x} = \frac{8(\ln(x))^7}{x}$

Next, let's differentiate $8 \cdot \ln(x)$ .
Since 8 is a constant, we can just multiply it by the derivative of $\ln(x)$ , which is $\frac{1}{x}$ .

This gives us: $8 \cdot \frac{1}{x} = \frac{8}{x}$

$\frac{dy}{dx} = \frac{8(\ln(x))^7 + 8}{x}$