Math  /  Calculus

Question14. y=ln(2x3+1)y=\ln \left(2 x^{3}+1\right).

Studdy Solution

STEP 1

Assumptions
1. The function given is y=ln(2x3+1) y = \ln(2x^3 + 1) .
2. We need to find the derivative dydx\frac{dy}{dx}.
3. The derivative of ln(u)\ln(u) with respect to xx is 1ududx\frac{1}{u} \cdot \frac{du}{dx}.

STEP 2

Identify the inner function u u in the expression y=ln(u) y = \ln(u) .
Let u=2x3+1 u = 2x^3 + 1 .

STEP 3

Differentiate the inner function u u with respect to x x .
dudx=ddx(2x3+1)\frac{du}{dx} = \frac{d}{dx}(2x^3 + 1).

STEP 4

Apply the power rule to differentiate 2x3 2x^3 .
The power rule states that ddx(xn)=nxn1\frac{d}{dx}(x^n) = nx^{n-1}.
ddx(2x3)=32x31=6x2\frac{d}{dx}(2x^3) = 3 \cdot 2x^{3-1} = 6x^2.

STEP 5

Differentiate the constant term in u u .
The derivative of a constant is zero.
ddx(1)=0\frac{d}{dx}(1) = 0.

STEP 6

Combine the derivatives from STEP_4 and STEP_5 to find dudx\frac{du}{dx}.
dudx=6x2+0=6x2\frac{du}{dx} = 6x^2 + 0 = 6x^2.

STEP 7

Use the chain rule to find dydx\frac{dy}{dx}.
The chain rule for y=ln(u) y = \ln(u) is dydx=1ududx\frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx}.

STEP 8

Substitute u=2x3+1 u = 2x^3 + 1 and dudx=6x2\frac{du}{dx} = 6x^2 into the chain rule formula.
dydx=12x3+16x2\frac{dy}{dx} = \frac{1}{2x^3 + 1} \cdot 6x^2.

STEP 9

Simplify the expression for dydx\frac{dy}{dx}.
dydx=6x22x3+1\frac{dy}{dx} = \frac{6x^2}{2x^3 + 1}.
The derivative of y=ln(2x3+1) y = \ln(2x^3 + 1) with respect to x x is dydx=6x22x3+1\frac{dy}{dx} = \frac{6x^2}{2x^3 + 1}.

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