Mathematics, 21.01.2020 10:31 StrangerMo

Find the derivative of (xlnx)/(1+lnx)

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Answer from: batista68

$\displaystyle \frac{dy}{dx} = 1 - \frac{\ln x}{\big( \ln (x) + 1 \big)^2}$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Product Rule]: $\displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle y = \frac{x \ln x}{1 + \ln x}$

Step 2: Differentiate

Derivative Rule [Quotient Rule]: $\displaystyle y' = \frac{(x \ln x)'(1 + \ln x) - (x \ln x)(1 + \ln x)'}{(1 + \ln x)^2}$ Derivative Rule [Product Rule]: $\displaystyle y' = \frac{[(x)' \ln x + x(\ln x)'](1 + \ln x) - (x \ln x)(1 + \ln x)'}{(1 + \ln x)^2}$ Logarithmic Differentiation [Derivative Properties]: $\displaystyle y' = \frac{[(x)' \ln x + 1](1 + \ln x) - \ln x}{(1 + \ln x)^2}$ Basic Power Rule: $\displaystyle y' = \frac{\big( \ln (x) + 1 \big) (1 + \ln x) - \ln x}{(1 + \ln x)^2}$ Simplify: $\displaystyle y' = \frac{(1 + \ln x)^2 - \ln x}{(1 + \ln x)^2}$ Rewrite: $\displaystyle y' = \frac{(1 + \ln x)^2}{(1 + \ln x)^2} - \frac{\ln x}{(1 + \ln x)^2}$ Simplify: $\displaystyle y' = 1 - \frac{\ln x}{(1 + \ln x)^2}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Answer from: Quest

x=12 units bd= 10 units

step-by-step explanation:

Answer from: Quest

step-by-step explanation:

each line from a vertex to the midpoint of the opposite side is called a median. the point where the medians intersect, point q, is the centroid.

the centroid of δabc is point q.

Lolol name the following segment or point. given: l, m, n are midpoints centroid of triangle abc p

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