Finding Tangent Equation for Curve $y=x^2$
To find the equation of the tangent line to the curve $y = x^2$ at the point $(1,1)$, we need the slope of the tangent at that point. The slope is given by the derivative of the function evaluated at the point.
Derivative Calculation
The curve is given by the equation:
$y = x^2$
Find the derivative of $y$ with respect to $x$ ($\frac{dy}{dx}$):
$\frac{dy}{dx} = \frac{d}{dx}(x^2)$
$\frac{dy}{dx} = 2x$
Slope at Point (1,1)
Evaluate the derivative at the point $(1,1)$ where $x=1$ to find the slope ($m$) of the tangent line:
$m = \frac{dy}{dx}\bigg|_{x=1} = 2(1)$
$m = 2$
Tangent Line Equation
Use the point-slope form of a linear equation, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point $(1,1)$ and $m$ is the slope $2$.
Substitute the values:
$y - 1 = 2(x - 1)$
Comparing with Options
The derived equation $y - 1 = 2(x - 1)$ matches Option 3 directly.
Important Questions from Functions Of Single Variable
- The gradient of $y = 3x^2 \sin(2x)$ at (0.2, 1) is __________ (rounded off to three decimal places).
- Let $ f(x) = x - [x] $, where $ x \ge 0 $ and $ [x] $ is the greatest integer not larger than x. Then $ f(x) $ is a
Consider the hyperbolic functions in Group – 1 and their definitions in Group - 2.
Group - 1 Group - 2 P $\tanh x$ I $\frac{e^x + e^{-x}}{e^x - e^{-x}}$ Q $\coth x$ II $\frac{2}{e^x + e^{-x}}$ R $\text{sech } x$ III $\frac{2}{e^x - e^{-x}}$ S $\text{cosech } x$ IV $\frac{e^x - e^{-x}}{e^x + e^{-x}}$ The correct combination is
- The figure which represents $y = \frac{\sin x}{x}$ for $x > 0$ (x in radians) is
- Consider the function $y = e^x$. The slope of this function at $x = 10$ is
