4.4 Solved examples of differentiation

  Basic Mathematics and Measurements

Question: Find the derivative of $y=x^3$ with respect to $x$.

Solution: Given, $$y = {x^3}$$ As, $$y = {x^n}$$$$\frac{{dy}}{{dx}} = n{x^{n - 1}}$$ So, $$y = {x^3}$$$$\frac{{dy}}{{dx}} = 3{x^{3 - 1}} = 3{x^2}$$

Question: $y = {x^2} + x + \frac{2}{x} + c$ Find $\frac{{dy}}{{dx}}$?

Solution: Given, $$y = {x^2} + x + \frac{2}{x} + c$$$$y = {x^2} + x + 2{x^{ - 1}} + c$$$$\frac{{dy}}{{dx}} = 2x + 1 - 2{x^{ - 2}}+0$$$$\frac{{dy}}{{dx}} = 2x + 1 - \frac{2}{{{x^2}}}$$

Question: FInd the derivative of $x\sin x$ wrt $x$?

Solution: Given, $$y = x\sin x$$ We will use the product rule for this kind of differentiation.
$$\frac{d}{{dx}}\left( {uv} \right) = u\frac{{dv}}{{dx}} + v\frac{{du}}{{dx}}$$Therefore, $$\frac{d}{{dx}}\left( {x\sin x} \right) = x\frac{{d\sin x}}{{dx}} + \sin x\frac{{dx}}{{dx}}$$$$\frac{{dy}}{{dx}} = x\cos x + \sin x$$

Question: Find the derivative of $\frac{{\sin x}}{x}$ wrt to $x$?

Solution: Given, $$y = \frac{{\sin x}}{x}$$ We will use the division rule for this kind of differentiation.
$$\frac{d}{{dx}}\left( {\frac{u}{v}} \right) = \frac{{v\frac{{du}}{{dx}} - u\frac{{dv}}{{dx}}}}{{{v^2}}}$$ So, $$\frac{d}{{dx}}\left( {\frac{{\sin x}}{x}} \right) = \frac{{x\frac{{d\sin x}}{{dx}} - \sin x\frac{{dx}}{{dx}}}}{{{x^2}}}$$$$\frac{{dy}}{{dx}} = \frac{{x\cos x - \sin x}}{{{x^2}}}$$

Question: Find the derivative of $y = {e^{{x^2} + 2}}$ wrt to $x$?

Solution: Given, $$y = {e^{{x^2} + 2}}$$ We will use chain rule for this kind of differentiation.

Let, $${x^2} + 2 = t \quad ...(i)$$ So, $$y = {e^t} \quad ...(ii)$$
We have to evaluate $\frac{{dy}}{{dx}}$. But now the $y$ is related to $t$ in the above equation.

So, we will relate $y$ with $t$ (equation $(ii)$) and $t$ with $x$ (equation $(ii)$) by using the chain rule.
$$\frac{{dy}}{{dx}} = \frac{{dy}}{{dt}} \times \frac{{dt}}{{dx}} \quad...(iii)$$
From equation $(i)$, $(ii)$ and $(iii)$ we can write,
$$\frac{{dy}}{{dx}} = \frac{{d\left( {{e^t}} \right)}}{{dt}} \times \frac{{d\left( {{x^2} + 2} \right)}}{{dx}}$$$$\frac{{dy}}{{dx}} = {e^t}\left( {2x} \right)$$$$\frac{{dy}}{{dx}} = 2x{e^{{x^2} + 2}}$$