Practice Questions (Step-by-Step Solutions)
Question 1
Find the derivative of [y] w.r.t [x]
[x = t^2], [y = \sin t]
Find [\dfrac{dy}{dx}].
Step-by-Step Solution:
- [\dfrac{dx}{dt} = 2t]
- [\dfrac{dy}{dt} = \cos t]
- [\dfrac{dy}{dx} = \dfrac{\cos t}{2t}]
Conclusion: Slope = [\dfrac{\cos t}{2t}]
Question 2
Find the derivative of [y] w.r.t [x]
[x = 3t + 2], [y = t^3]
Solution:
- [\dfrac{dx}{dt} = 3]
- [\dfrac{dy}{dt} = 3t^2]
- [\dfrac{dy}{dx} = \dfrac{3t^2}{3} = t^2]
Question 3
Find the derivative of [y] w.r.t [x]
[x = \ln t], [y = t^2]
- [\dfrac{dx}{dt} = \dfrac{1}{t}]
- [\dfrac{dy}{dt} = 2t]
- [\dfrac{dy}{dx} = \dfrac{2t}{1/t} = 2t^2]
Question 4
Find the derivative of [y] w.r.t [x]
[x = \sec t], [y = \tan t]
- [\dfrac{dx}{dt} = \sec t\tan t]
- [\dfrac{dy}{dt} = \sec^2 t]
- [\dfrac{dy}{dx} = \dfrac{\sec^2 t}{\sec t\tan t} = \csc t \sec t]
Question 5
Find the derivative of [y] w.r.t [x]
[x = t + \dfrac{1}{t}], [y = t – \dfrac{1}{t}]
- [\dfrac{dx}{dt} = 1 – \dfrac{1}{t^2}]
- [\dfrac{dy}{dt} = 1 + \dfrac{1}{t^2}]
- [\dfrac{dy}{dx} = \dfrac{1 + \dfrac{1}{t^2}}{1 – \dfrac{1}{t^2}}]
Question 6
Find the derivative of [y] w.r.t [x]
[x = 4\cos t], [y = 5\sin t]
- [\dfrac{dx}{dt} = -4\sin t]
- [\dfrac{dy}{dt} = 5\cos t]
- [\dfrac{dy}{dx} = \dfrac{5\cos t}{-4\sin t} = -\dfrac{5}{4} \cot t]
Question 7
Find the derivative of [y] w.r.t [x]
[x = e^{-t}], [y = \ln t]
- [\dfrac{dx}{dt} = -e^{-t}]
- [\dfrac{dy}{dt} = \dfrac{1}{t}]
- [\dfrac{dy}{dx} = -\dfrac{1}{t e^{-t}} = -\dfrac{e^t}{t}]
Question 8
Find the derivative of [y] w.r.t [x]
[x = \tan t], [y = \sec t]
- [\dfrac{dx}{dt} = \sec^2 t]
- [\dfrac{dy}{dt} = \sec t \tan t]
- [\dfrac{dy}{dx} = \dfrac{\sec t \tan t}{\sec^2 t} = \sin t]
Question 9
Find the derivative of [y] w.r.t [x]
[x = t^2 + 1], [y = \ln(t^2 + 1)]
- [\dfrac{dx}{dt} = 2t]
- [\dfrac{dy}{dt} = \dfrac{2t}{t^2 + 1}]
- [\dfrac{dy}{dx} = \dfrac{2t}{2t(t^2 + 1)} = \dfrac{1}{t^2 + 1}]
Question 10
Find the derivative of [y] w.r.t [x]
[x = \sinh t], [y = \cosh t]
- [\dfrac{dx}{dt} = \cosh t]
- [\dfrac{dy}{dt} = \sinh t]
- [\dfrac{dy}{dx} = \dfrac{\sinh t}{\cosh t} = \tanh t]