Question

Find \(\frac{{dy}}{{dx}}\) and \(\frac{{dy}}{{dt}}\).

\(y = t{x^2} + {t^3}x\)

Short answer

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

Step-by-step solution

Different rules of differentiation

The given function is \(y = t{x^2} + {t^3}x\).

The sum rule is:\(\frac{d}{{dx}}\left( {g\left( x \right) + f\left( x \right)} \right) = \frac{d}{{dx}}\left( {g\left( x \right)} \right) + \frac{d}{{dx}}\left( {f\left( x \right)} \right)\)

And, the power rule is: \(\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\)

Evaluating Derivative using all these rules

Differentiating with respect to\(x\)taking\(t\)as a constant:

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

Hence, the required value is \(\frac{{dy}}{{dx}} = 2xt + {t^3}\).

Evaluating Derivative for \(t\)

Differentiating with respect to\(t\)taking\(x\)as a constant:

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

Hence, the required value is \(\frac{{dy}}{{dt}} = {x^2} + 3{t^2}x\).