Math  /  Calculus

QuestionLet kk be a constant. Compute ddx[ln(kx)]\frac{d}{d x}[\ln (k x)] in two ways. a) Using the Chain Rule, first decompose y=ln(kx)y=\ln (k x) into an outside and inside function Outside function (in terms of uu ): y=y= \square Inside function (in terms of xx ): u=u= \square .
Then find the derivative, ddx[ln(kx)]=\frac{d}{d x}[\ln (k x)]= \square (simplify your answer). b) Using a law of logarithms to simplify first: ln(kx)=\ln (k x)= \square +vundefined+\widehat{v} lnx\ln x. (Fill in the blanks to make this a true statement.) Now take the derivative of the simplified function: dydx=\frac{d y}{d x}= \square Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of ln(kx)\ln(kx) in two different ways: first using the chain rule, and then by simplifying the expression with logarithm rules before taking the derivative. Watch out! Remember that the derivative of ln(x)\ln(x) is 1/x1/x, not 1/kx1/kx when dealing with ln(kx)\ln(kx).
Also, be careful with the chain rule and logarithm rules!

STEP 2

1. Chain Rule Method
2. Logarithm Rule Method

STEP 3

Alright, let's **define** our functions!
Our **outside function** is y=ln(u)y = \ln(u), where uu is our **inside function** u=kxu = kx.
Think of it like a delicious turducken – kxkx is the duck inside the chicken, which is inside the turkey, which is like our logarithm!

STEP 4

Now, let's find the **derivatives**!
The derivative of the **outside function** y=ln(u)y = \ln(u) with respect to uu is dydu=1u\frac{dy}{du} = \frac{1}{u}.
The derivative of the **inside function** u=kxu = kx with respect to xx is dudx=k\frac{du}{dx} = k.

STEP 5

Time for the **chain rule**!
Remember, it's like multiplying the rates.
We multiply the derivative of the outside function by the derivative of the inside function: dydx=dydududx=1uk=ku.\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot k = \frac{k}{u}.

STEP 6

Since u=kxu = kx, we **substitute** that back in, giving us dydx=kkx\frac{dy}{dx} = \frac{k}{kx}.
Now, we can **simplify** by dividing both the numerator and the denominator by kk, which is the same as multiplying by 1/k1/k=1\frac{1/k}{1/k} = 1.
This gives us the **final result**: dydx=1x\frac{dy}{dx} = \frac{1}{x}.

STEP 7

Using our logarithm rule, we can **rewrite** ln(kx)\ln(kx) as ln(k)+ln(x)\ln(k) + \ln(x).
It's like breaking down a big log into smaller, manageable pieces!

STEP 8

Now, we take the **derivative** with respect to xx: ddx(ln(k)+ln(x))=ddx(ln(k))+ddx(ln(x)).\frac{d}{dx}(\ln(k) + \ln(x)) = \frac{d}{dx}(\ln(k)) + \frac{d}{dx}(\ln(x)). Since ln(k)\ln(k) is a constant, its derivative is **zero**.
The derivative of ln(x)\ln(x) is 1x\frac{1}{x}.
So, adding these together, we get dydx=0+1x=1x\frac{dy}{dx} = 0 + \frac{1}{x} = \frac{1}{x}.

STEP 9

Using both the chain rule and the logarithm rule, we found that the derivative of ln(kx)\ln(kx) with respect to xx is 1x\frac{1}{x}.

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