• Let \(y\) be a function of \(x\) i.e., \(y=f(x)\) and \(\Delta y\) be the change in \(y\) corresponding to a small change \((\Delta x)\) in \(x\).
• Then, \(\dfrac{\Delta y}{\Delta x}\) represents the average rate of change of \(y\) with respect to \(x\) as \(x\) changes  from \(x\) to \(x + \Delta x\).
• As \(\Delta x \to 0\), the limiting  value of this average rate of change of \(y\) with respect to \(x\) in the interval  \([x, x + \Delta x]\) becomes the instantaneous rate of change of \(y\) with respect to \(x\) i.e., \(\left (\dfrac{dy}{dx}\right)\)
• Thus, \(\lim \limits_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x}\) = Instantaneous rate of change of \(y\) with respect to \(x= \dfrac{dy}{dx}\)

             \(\Rightarrow\) \( \dfrac{dy}{dx}\) = Instantaneous rate of change of \(y\) with respect to \(x\)

                    \(\left[\because\, \lim \limits_{\Delta x \to 0} \dfrac{\Delta y}{\Delta x}= \dfrac{dy}{dx}\right]\)

• Hence, to determine the instantaneous rate of change of one quantity with respect to another quantity, calculate its derivative.
• Suppose the position \((x)\) of a particle is given as the function of time \((t)\) while it is moving along a straight line.

            Then, average velocity = \(\dfrac{\Delta x}{\Delta t}\) 

and, instantaneous velocity = \(\dfrac{d x}{d t}\)

• To determine the instantaneous acceleration, take the derivative of instantaneous velocity \(\left (\dfrac{dx}{dt}\right)\) i.e., \(a_{inst} = \dfrac{d}{dt}\left(\dfrac{dx}{dt}\right)\)

            \(a_{inst} = \dfrac{d}{dt}\left(v_{inst}\right)\) 

            \(a_{inst} = \dfrac{d \,v_{inst}}{dt}\)              

• Let \(y\) be a function of \(x\), as given

           \(y= f(x) = x^n\)

where 'n' is a real number

            Then, the derivative of the function \(y= f(x)\) will be  

           \(\dfrac{dy}{dx} = n x^{n-1}\)

• Let \(y\) be a function of \(x\) and \(x\) be a function of t.

          Then, by chain rule,

                \(\dfrac{dy}{dt} =\dfrac{dy}{dx} . \dfrac{dx}{dt} \)       

• \(\dfrac{d}{dx} (sin \, x) = cos \,x\)
• \(\dfrac{d}{dx} (cos\, x) = -sin \,x\)
• \(\dfrac{d}{dx} (tan\, x) = sec^2 \,x\)
• \(\dfrac{d}{dx} (cot\, x) = -cosec^2\,x\)
• \(\dfrac{d}{dx} (sec\, x) = sec \,x \, tan \,x\)
• \(\dfrac{d}{dx} (cosec\, x) = -cosec \,x \, cot \,x\)

• \(\dfrac{d}{dx}(e^x) = e^x\)
• \(\dfrac{d}{dx}(\ell n\,x) = \dfrac{1}{x}\)

• If a tangent to the curve \(y= f(x)\) makes an angle \(\theta\) with \(x - axis\) in the positive direction, then \(\dfrac{dy}{dx}\)= slope of tangent = \(tan \,\theta\)
• The slope of the tangent to the curve \(y= f(x)\) at the point \((x_1 , y_1)\) is given by \(\left[\dfrac{dy}{dx}\right]_{(x_1,y_1)}\) .

• In the given \(y\) versus \(x \) graph, curve \(y = f(x)\) is shown.
• The point 'A' represents the highest point of the curve. So, it is the point of local maxima.
• Point 'B' represents the lowest point of the curve. Hence, it is the point of local minima.
• The tangents drawn at these two points have zero slope.
• So, it can be concluded that at maxima and minima,

\(\dfrac{dy}{dx} = 0\)

• For one - dimensional motion,

           \(v=\dfrac{dx}{dt}\)

         When velocity (\(v\)) becomes zero, displacement \((x)\) is said to be maximum or minimum.

• Similarly,

            \(a = \dfrac{dv}{dt}\)  

         Here, when acceleration becomes zero, velocity is said to be maximum or minimum.

• To check whether the values are maximum or minimum, the double derivatives are to be found.
• Thus, condition of maxima is

             \(\dfrac{dy}{dx} = 0\) 

      and, \(\dfrac{d^2y}{dx^2} < 0\)

• Condition of minima is

         \(\dfrac{dy}{dx} = 0\)

         and, \(\dfrac{d^2y}{dx^2} > 0\)

• If  \(y= uv\)  is a product of two functions \(u\) and \(v\), where \(u= f(x) \) and \(v= g(x)\).

• Then, derivative of \(y= uv\) is given as

  \(\dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}\)

• Let \(y\) be the function of \(x\) i.e., \(y= f(x)\).
• When we differentiate \(y\) with respect to \(x\) two times successively, then this process is called double differentiation.
•  It is represented as \(\dfrac{d^2y}{dx^2}\)

           \(\dfrac{d^2y}{dx^2} = \dfrac{d}{dx} \left(\dfrac{dy}{dx}\right)\)

• For example - 

            Instantaneous velocity, \(v= \dfrac{dx}{dt}\) and instantaneous acceleration, \(a= \dfrac{dv}{dt}\)

             \(\Rightarrow a = \dfrac{d}{dt} \left(\dfrac{dx}{dt}\right)\)

             \(\Rightarrow a = \dfrac{d^2x}{dt^2}\)