Let $y=\frac{1-2 x}{3 x^{2}}$.
What is the value of $\frac{d y}{d x}$ at $x=1$ ?

Question

Let $y=\frac{1-2 x}{3 x^{2}}$.
What is the value of $\frac{d y}{d x}$ at $x=1$ ?

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Given the function $y=\frac{1-2x}{3x^2}$, we need to find the derivative $\frac{dy}{dx}$. ‖ ‖ step 2 ⋮ From the asksia-ll calculation list, the derivative of the function with respect to $x$ is $\frac{dy}{dx} = \frac{2(x-1)}{3x^3}$. ‖ ‖ step 3 ⋮ To find the value of $\frac{dy}{dx}$ at $x=1$, we substitute $x=1$ into the derivative. ‖ ‖ step 4 ⋮ Substituting $x=1$ gives $\frac{dy}{dx} = \frac{2(1-1)}{3(1)^3} = \frac{2(0)}{3} = 0$. ‖ ∻Answer∻ ⚹ $\frac{dy}{dx} = 0$ at $x=1$ ⚹ ∻Key Concept∻ ⚹Differentiation and evaluation at a point⚹ ∻Explanation∻ ⚹The value of the derivative of a function at a specific point is found by substituting the point into the derivative expression.⚹◊What is the derivative of $y=\frac{1-2x}{3x^2}$ with respect to $x$?⍭ Generate me a similar question◊

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