Question

How do you find the derivative of \[x{y^2}\]?

Answer 1

Hint:Here, we are required to find the derivative of the given product of two variables. Thus, we will use the product rule and substitute the first variable as the first term and the second variable as the second term. Differentiating with respect to \[x\] and using the product rule will help us to find the required derivative of the given function.

Formula Used:
According to the product rule of two terms of a given function,
\[\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I\dfrac{d}{{dx}}\left( {II} \right) + II\dfrac{d}{{dx}}\left( I \right)\]

Complete step-by-step answer:
In order to find the derivative of \[x{y^2}\], we will use the product rule of derivatives.
According to the product rule,
\[\dfrac{d}{{dx}}\left( {I \cdot II} \right) = I\dfrac{d}{{dx}}\left( {II} \right) + II\dfrac{d}{{dx}}\left( I \right)\]
Hence, using product rule in \[x{y^2}\], we get,
\[\dfrac{d}{{dx}}\left( {x \cdot {y^2}} \right) = x\dfrac{d}{{dx}}\left( {{y^2}} \right) + {y^2}\dfrac{d}{{dx}}\left( x \right)\]
By power and chain rule, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}}\left( {x \cdot {y^2}} \right) = 2xy\dfrac{{dy}}{{dx}} + {y^2}\]

Therefore, the derivative of \[x{y^2}\] is \[2xy\dfrac{{dy}}{{dx}} + {y^2}\]

Also, if we would have done the differentiation with respect to \[t\] using the product rule, then the derivative would be:
\[\dfrac{d}{{dt}}\left( {x \cdot {y^2}} \right) = x\dfrac{d}{{dt}}\left( {{y^2}} \right) + {y^2}\dfrac{d}{{dt}}\left( x \right)\]
\[ \Rightarrow \dfrac{d}{{dt}}\left( {x \cdot {y^2}} \right) = 2xy\dfrac{{dy}}{{dt}} + {y^2}\dfrac{{dx}}{{dt}}\]
Hence, this would have been our answer and the derivative of \[x\] would have not been equal to 1.
Thus, this is the required answer.

Note: In mathematics, the rate of change of a function with respect to a variable is known as a derivative. Integration is the opposite of integration and therefore known as the antiderivative. According to the product rule in derivative, if the two parts of a function are being multiplied together then, we write that product two times, being added together and then, we find the derivative of the first term in the first product and the derivative of the second term in the second product.