Question

The equation of the tangent to the curve \(y=x^3\) at \((1,1)\) is

• A. \(3x-y+2=0\)
• B. \(x-10y-50=0\)
• C. \(3x-y-2=0\)
• D. \(x-10y+50=0\)

Hint:

To find tangent equation, first find slope using derivative, then use \(y-y_1=m(x-x_1)\).

Answer 1

The Correct Option is C

We are given the curve: \[ y=x^3. \] We need the equation of tangent at the point: \[ (1,1). \] First find the slope of tangent. Differentiate: \[ \frac{dy}{dx}=\frac{d}{dx}(x^3). \] \[ \frac{dy}{dx}=3x^2. \] At \(x=1\): \[ \frac{dy}{dx}=3(1)^2. \] \[ \frac{dy}{dx}=3. \] So the slope of the tangent is \[ m=3. \] Now use point-slope form: \[ y-y_1=m(x-x_1). \] Here, \[ (x_1,y_1)=(1,1) \] and \[ m=3. \] Therefore, \[ y-1=3(x-1). \] Expand: \[ y-1=3x-3. \] Bring all terms to one side: \[ 3x-y-2=0. \] Hence, the equation of the tangent is \[ 3x-y-2=0. \]