Accedmychevron_right11thchevron_rightmathchevron_rightDifferentiationchevron_rightExercise 13.1

Questions

  1. Question 1

    Find by definition, the derivatives w.r.t. ‘x’ of the following functions defined as:

  2. Question 2

    Find dydx\dfrac{dy}{dx} from first principle and find gradient of the curve at the given point:

  3. Question 4

    Find from first principle, the derivatives of the following expressions w.r.t. their respective independent variables:

  4. Question 5

    Find the gradient and equation of the tangent line to y=3x24x+1y = 3x^2 - 4x + 1 at x=2x = 2.

  5. Question 6

    For the function f(x)=2x3+xf(x) = 2x^3 + x, calculate the equation of the tangent line at x=1x = -1.

  6. Question 7

    Find the coordinates of the point of tangency and the equation of the tangent line for f(x)=x32x+1f(x) = x^3 - 2x + 1 at x=1x = 1.

  7. Question 8

    Find the gradient of the curve f(x)=3x2+2xf(x) = 3x^2 + 2x at x=1x = 1.

  8. Question 9

    Find the gradient and an equation of tangent line to the graph of f(x)=xf(x) = \sqrt{x} at x=9x = 9.

  9. Question 10

    The position of a car after tt hours is given by: s(t)=2t33t2+ts(t) = 2t^3 - 3t^2 + t (in kilometres)

  10. Question 11

    A stone is thrown upwards and its height after tt seconds is given by: s(t)=16t2+32t+10s(t) = -16t^2 + 32t + 10 (in feet). Find the instantaneous velocity at t=1t = 1.

  11. Question 12

    The outdoor temperature (in °C) over time is modeled by: T(t)=t2+12t+10T(t) = -t^2 + 12t + 10, where tt is the time in hours. Find the instantaneous rate of change at t=2t = 2.