Find dy/dx y=x natural log of x-x
The problem presented here is asking for the derivative of the function y with respect to x, where y is defined as the difference between x times the natural logarithm of x (denoted as ln(x)) and x itself. To clarify, the function can be written as y = x * ln(x) - x. The task is to perform differentiation on this equation to determine the rate of change of y with respect to x, i.e., dy/dx. This question involves applying the rules of calculus, particularly the product rule, as the function consists of the product of two functions of x (x and ln(x)), and then simplifying the expression to find the derivative.
$y = x ln \left(\right. x \left.\right) - x$
Answer
Solution:
Step:1
Take the derivative of both sides with respect to $x$: $\frac{d}{dx}(y) = \frac{d}{dx}(x \ln(x) - x)$.
Step:2
The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$.
Step:3
Proceed to differentiate the expression on the right-hand side.
Step:3.1
Apply the Sum Rule to find the derivative of the sum and difference: $\frac{d}{dx}(x \ln(x)) + \frac{d}{dx}(-x)$.
Step:3.2
Compute $\frac{d}{dx}(x \ln(x))$.
Step:3.2.1
Use the Product Rule, which is given by $\frac{d}{dx}(uv) = u \frac{d}{dx}(v) + v \frac{d}{dx}(u)$, where $u = x$ and $v = \ln(x)$.
Step:3.2.2
The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$, thus we have $x \cdot \frac{1}{x} + \ln(x) \cdot \frac{d}{dx}(x) + \frac{d}{dx}(-x)$.
Step:3.2.3
Apply the Power Rule, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$, where $n = 1$.
Step:3.2.4
Simplify the expression by combining $x$ and $\frac{1}{x}$.
Step:3.2.5
Eliminate the common factors.
Step:3.2.5.1
Cancel out the common $x$ terms.
Step:3.2.5.2
Simplify the expression to $1 + \ln(x) + \frac{d}{dx}(-x)$.
Step:3.2.6
Multiply $\ln(x)$ by $1$ to get $1 + \ln(x) + \frac{d}{dx}(-x)$.
Step:3.3
Find the derivative of $-x$.
Step:3.3.1
Since $-1$ is a constant, the derivative of $-x$ is $-1 \cdot \frac{d}{dx}(x)$.
Step:3.3.2
Use the Power Rule with $n = 1$.
Step:3.3.3
Multiply $-1$ by $1$ to get $1 + \ln(x) - 1$.
Step:3.4
Combine like terms.
Step:3.4.1
Subtract $1$ from $1$ to get $0 + \ln(x)$.
Step:3.4.2
Combine $0$ and $\ln(x)$ to get $\ln(x)$.
Step:4
Write the derivative of $y$ as equal to the simplified right-hand side: $\frac{dy}{dx} = \ln(x)$.
Step:5
Substitute $\frac{dy}{dx}$ for $y$ in the final expression: $\frac{dy}{dx} = \ln(x)$.
Knowledge Notes:
Derivative: The derivative of a function measures how the function value changes as its input changes. The notation $\frac{d}{dx}$ is used to denote the derivative with respect to $x$.
Sum Rule: The derivative of a sum of functions is the sum of their derivatives.
Product Rule: For two functions $u(x)$ and $v(x)$, the derivative of their product $uv$ is given by $u'v + uv'$.
Power Rule: The derivative of $x^n$ with respect to $x$ is $nx^{n-1}$.
Natural Logarithm: The natural logarithm function, denoted as $\ln(x)$, is the inverse of the exponential function $e^x$. The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$.
Combining Like Terms: Simplifying expressions by adding or subtracting like terms, which are terms that have the same variables raised to the same power.
Simplification: The process of reducing an expression to its simplest form by performing arithmetic operations and combining like terms.
