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

Apply Product Rule: Using the product rule, which states that (d/dx)[u*v] = u'v + uv', where u = x^(4) and v = ln(x). Differentiate u with respect to x to get u' = (d/dx)x^(4) = 4x^(3).Differentiate u: Now, differentiate v with respect to x to get v' = (d/dx)ln(x) = 1/x.Differentiate v: Plug u', u, v', and v into the product rule formula: (4x^(3))*ln(x) + (x^(4))*(1/x).Apply Product Rule Formula: Simplify the expression: 4x^(3)ln(x) + x^(3).

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Let
y=x^(4)ln(x).
Find
(dy)/(dx).
Choose 1 answer:
(A)
4x^(2)
(B)
x^(3)(4ln(x)+1)
(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) $4 x^{2}$ $\newline$ (B) $x^{3}(4 \ln (x)+1)$ $\newline$ (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)

4 x^{2}

\newline

(B)

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

\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, which states that $(\frac{d}{dx})[u*v] = u'v + uv'$ , where $u = x^{4}$ and $v = \ln(x)$ . Differentiate $u$ with respect to $x$ to get $u' = (\frac{d}{dx})x^{4} = 4x^{3}$ .
Differentiate $u$ : Now, differentiate $v$ with respect to $x$ to get $v' = \frac{d}{dx}\ln(x) = \frac{1}{x}$ .
Differentiate $v$ : Plug $u'$ , $u$ , $v'$ , and $v$ into the product rule formula: $(4x^{3})\cdot\ln(x) + (x^{4})\cdot(\frac{1}{x})$ .
Apply Product Rule Formula: Simplify the expression: $4x^{3}\ln(x) + x^{3}.$