Math  /  Calculus

QuestionDifferentiate the following function. y=4ln(3x)xy=\frac{4 \ln (3 x)}{\sqrt{x}} ddx(4ln(3x)x)=\frac{d}{d x}\left(\frac{4 \ln (3 x)}{\sqrt{x}}\right)=

Studdy Solution

STEP 1

1. We are given the function y=4ln(3x)x y = \frac{4 \ln(3x)}{\sqrt{x}} .
2. We need to find the derivative of this function with respect to x x .

STEP 2

1. Rewrite the function in a form that is easier to differentiate.
2. Apply the quotient rule to differentiate the function.
3. Simplify the derivative expression.

STEP 3

Rewrite the function y=4ln(3x)x y = \frac{4 \ln(3x)}{\sqrt{x}} in terms of exponents:
y=4ln(3x)x1/2 y = 4 \ln(3x) \cdot x^{-1/2}

STEP 4

Apply the product rule to differentiate y=4ln(3x)x1/2 y = 4 \ln(3x) \cdot x^{-1/2} .
The product rule states that if u(x) u(x) and v(x) v(x) are functions of x x , then:
ddx[u(x)v(x)]=u(x)v(x)+u(x)v(x) \frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
Let u(x)=4ln(3x) u(x) = 4 \ln(3x) and v(x)=x1/2 v(x) = x^{-1/2} .

STEP 5

Differentiate u(x)=4ln(3x) u(x) = 4 \ln(3x) :
u(x)=43x3=4x u'(x) = \frac{4}{3x} \cdot 3 = \frac{4}{x}

STEP 6

Differentiate v(x)=x1/2 v(x) = x^{-1/2} :
v(x)=12x3/2 v'(x) = -\frac{1}{2}x^{-3/2}

STEP 7

Apply the product rule using the derivatives found:
ddx(4ln(3x)x1/2)=(4x)x1/2+(4ln(3x))(12x3/2) \frac{d}{dx}\left(4 \ln(3x) \cdot x^{-1/2}\right) = \left(\frac{4}{x}\right)x^{-1/2} + \left(4 \ln(3x)\right)\left(-\frac{1}{2}x^{-3/2}\right)

STEP 8

Simplify the expression:
=4x3/22ln(3x)x3/2 = \frac{4}{x^{3/2}} - 2 \ln(3x) \cdot x^{-3/2}
=42ln(3x)x3/2 = \frac{4 - 2 \ln(3x)}{x^{3/2}}
The derivative of the function is:
ddx(4ln(3x)x)=42ln(3x)x3/2 \frac{d}{dx}\left(\frac{4 \ln(3x)}{\sqrt{x}}\right) = \frac{4 - 2 \ln(3x)}{x^{3/2}}

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