Let y=x^(4)ln(x). Find (dy)/(dx). Choose 1 answer: (A) x^(3)(4ln(x)+1) (B) 4x^(2) (C) 4x^(3)+(1)/(x) (D) 4x^(3)(x^(4)+ln(x))

Apply Product Rule: Using the product rule: (d(uv))/(dx) = u(dv)/(dx) + v(du)/(dx), where u = x^(4) and v = ln(x).Find du/dx: First, find the derivative of u = x^(4), which is (du)/(dx) = 4x^(3).Find dv/dx: Now, find the derivative of v = ln(x), which is (dv)/(dx) = 1/x.Apply Product Rule: Apply the product rule: (dy)/(dx) = x^(4) * (1/x) + ln(x) * 4x^(3).Simplify Expression: Simplify the expression: (dy)/(dx) = x^(3) + 4x^(3)ln(x). Oops, there's a mistake here. We need to multiply x^(3) by 4 in the second term.

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Let
y=x^(4)ln(x).
Find
(dy)/(dx).
Choose 1 answer:
(A)
x^(3)(4ln(x)+1)
(B)
4x^(2)
(C)
4x^(3)+(1)/(x)
(D)
4x^(3)(x^(4)+ln(x))

Let $y=x^{4} \ln (x)$ . $\newline$ Find $\frac{d y}{d x}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $x^{3}(4 \ln (x)+1)$ $\newline$ (B) $4 x^{2}$ $\newline$ (C) $4 x^{3}+\frac{1}{x}$ $\newline$ (D) $4 x^{3}\left(x^{4}+\ln (x)\right)$

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Math Problems

Algebra 1

Multiplication with rational exponents

Full solution

Q. Let

y=x^{4} \ln (x)

\newline

Find

\frac{d y}{d x}

\newline

Choose

1

answer:

\newline

(A)

x^{3}(4 \ln (x)+1)

\newline

(B)

4 x^{2}

\newline

(C)

4 x^{3}+\frac{1}{x}

\newline

(D)

4 x^{3}\left(x^{4}+\ln (x)\right)

Apply Product Rule: Using the product rule: $(d(uv))/(dx) = u(dv)/(dx) + v(du)/(dx)$ , where $u = x^{4}$ and $v = \ln(x)$ .
Find $\frac{du}{dx}$ : First, find the derivative of $u = x^{4}$ , which is $\frac{du}{dx} = 4x^{3}$ .
Find $\frac{dv}{dx}$ : Now, find the derivative of $v = \ln(x)$ , which is $\left(\frac{dv}{dx}\right) = \frac{1}{x}$ .
Apply Product Rule: Apply the product rule: $(\frac{dy}{dx}) = x^{4} \cdot (\frac{1}{x}) + \ln(x) \cdot 4x^{3}$ .
Simplify Expression: Simplify the expression: $(\frac{dy}{dx}) = x^{3} + 4x^{3}\ln(x)$ . Oops, there's a mistake here. We need to multiply $x^{3}$ by $4$ in the second term.