Question

Find \(dy/dx\) and \(dy/dt\)?

36. \(y = \frac{t}{{{x^2}}} + \frac{x}{t}\)

Short answer

The values are \(\frac{{dy}}{{dx}} = \frac{{ - 2t}}{{{x^3}}} + \frac{1}{t}\) and \(\frac{{dy}}{{dt}} = \frac{1}{{{x^2}}} - \frac{x}{{{t^2}}}\).

Step-by-step solution

Sum rule of differentiation

Consider \(t\) as constant and \(x\) as variable to find the derivative \(\frac{{dy}}{{dx}}\).

Use sum rule of differentiation which implies that \(\frac{d}{{dx}}\left( {f\left( x \right) + g\left( x \right)} \right) = \frac{d}{{dx}}f\left( x \right) + \frac{d}{{dx}}g\left( x \right)\).

\(\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{t}{{{x^2}}} + \frac{x}{t}} \right)\\ &= \frac{d}{{dx}}\left( {\frac{t}{{{x^2}}}} \right) + \frac{d}{{dx}}\left( {\frac{x}{t}} \right)\end{align}\)

Constant multiple rule of differentiation

The constant multiple rule of differentiation implies that \(\frac{d}{{dx}}\left( {cf\left( x \right)} \right) = c\frac{d}{{dx}}f\left( x \right)\), where \(c\) is a constant.

Apply constant multiple rule.

\(\begin{align}\frac{{dy}}{{dx}} &= \frac{d}{{dx}}\left( {\frac{t}{{{x^2}}}} \right) + \frac{d}{{dx}}\left( {\frac{x}{t}} \right)\\ &= t\frac{d}{{dx}}\left( {{x^{ - 2}}} \right) + \frac{1}{t}\frac{d}{{dx}}\left( x \right)\end{align}\)

Power rule of differentiation

The power rule of differentiation implies that \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\), where \(n\) is any real number.

Apply power rule and simplify.

\(\begin{align}\frac{{dy}}{{dx}} &= t\left( { - 2} \right){x^{ - 2 - 1}} + \frac{1}{t}\\ &= \frac{{ - 2t}}{{{x^3}}} + \frac{1}{t}\end{align}\)

Thus, \(\frac{{dy}}{{dx}} = \frac{{ - 2t}}{{{x^3}}} + \frac{1}{t}\).

Sum rule of differentiation

Consider \(x\) as constant and \(t\) as variable to find the derivative \(\frac{{dy}}{{dt}}\).

Apply sum rule of differentiation

\(\begin{align}\frac{{dy}}{{dt}} &= \frac{d}{{dt}}\left( {\frac{t}{{{x^2}}} + \frac{x}{t}} \right)\\ &= \frac{d}{{dt}}\left( {\frac{t}{{{x^2}}}} \right) + \frac{d}{{dt}}\left( {\frac{x}{t}} \right)\end{align}\)

Constant multiple rule of differentiation

Apply constant multiple rule.

\(\begin{align}\frac{{dy}}{{dt}} &= \frac{d}{{dt}}\left( {\frac{t}{{{x^2}}}} \right) + \frac{d}{{dt}}\left( {\frac{x}{t}} \right)\\ &= \frac{1}{{{x^2}}}\frac{d}{{dt}}\left( t \right) + x\frac{d}{{dt}}\left( {\frac{1}{t}} \right)\\ &= \frac{1}{{{x^2}}}\frac{d}{{dt}}\left( t \right) + x\frac{d}{{dt}}\left( {{t^{ - 1}}} \right)\end{align}\)

Power rule of differentiation

Apply product rule and simplify.

\(\begin{align}\frac{{dy}}{{dt}} &= \frac{1}{{{x^2}}}\left( 1 \right) + x\left( { - 1{t^{ - 1 - 1}}} \right)\\ &= \frac{1}{{{x^2}}} - \frac{x}{{{t^2}}}\end{align}\)

Thus, \(\frac{{dy}}{{dt}} = \frac{1}{{{x^2}}} - \frac{x}{{{t^2}}}\).