
What is the derivative of ${x^3}$ with respect to ${x^2}$ ?
(A) $3{x^2}$
(B) $\dfrac{{3x}}{2}$
(C) $x$
(D) $\dfrac{3}{2}$
Answer
490.5k+ views
Hint: In the given problem, we are required to differentiate the function ${x^3}$ with respect to ${x^2}$. Since we cannot differentiate the function with respect to ${x^2}$, we will use the chain rule of differentiation. Using the chain rule of differentiation, we will first find the derivative of ${x^3}$ and ${x^2}$ with respect to x and then divide both the expressions to get to the required answer. Power rule of differentiation must be remembered in order to solve the problem.
Complete step-by-step solution:
In the given question, we have to find the derivative of ${x^3}$ with respect to ${x^2}$.
So, we have, $\dfrac{{d\left( {{x^3}} \right)}}{{d\left( {{x^2}} \right)}}$.
So, first we evaluate the derivatives of both the functions ${x^3}$ and ${x^2}$ with respect to x.
So, we get the derivative of ${x^3}$ with respect to x as $\dfrac{{d\left( {{x^3}} \right)}}{{dx}}$.
Now, using the power rule of differentiation $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{{d\left( {{x^3}} \right)}}{{dx}} = 3{x^2}$
Now, we also find the derivative of ${x^2}$ with respect to x. Hence, we get, $\dfrac{{d\left( {{x^2}} \right)}}{{dx}}$.
Using the power rule of differentiation $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2x$
Now, we divide both the equations to find the derivative of ${x^3}$ with respect to ${x^2}$.
So, we get,
$ \Rightarrow \dfrac{{\dfrac{{d\left( {{x^3}} \right)}}{{dx}}}}{{\dfrac{{d\left( {{x^2}} \right)}}{{dx}}}} = \dfrac{{3{x^2}}}{{2x}}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow \dfrac{{d\left( {{x^3}} \right)}}{{d\left( {{x^2}} \right)}} = \dfrac{{3x}}{2}$
So, the derivative of ${x^3}$ with respect to ${x^2}$ is $\dfrac{{3x}}{2}$.
Therefore, option (A) is the correct answer to the problem.
Note: Derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite function and examining the behaviour of function layer by layer. We must take care of the calculations while solving such questions.
Complete step-by-step solution:
In the given question, we have to find the derivative of ${x^3}$ with respect to ${x^2}$.
So, we have, $\dfrac{{d\left( {{x^3}} \right)}}{{d\left( {{x^2}} \right)}}$.
So, first we evaluate the derivatives of both the functions ${x^3}$ and ${x^2}$ with respect to x.
So, we get the derivative of ${x^3}$ with respect to x as $\dfrac{{d\left( {{x^3}} \right)}}{{dx}}$.
Now, using the power rule of differentiation $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{{d\left( {{x^3}} \right)}}{{dx}} = 3{x^2}$
Now, we also find the derivative of ${x^2}$ with respect to x. Hence, we get, $\dfrac{{d\left( {{x^2}} \right)}}{{dx}}$.
Using the power rule of differentiation $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, we get,
$ \Rightarrow \dfrac{{d\left( {{x^2}} \right)}}{{dx}} = 2x$
Now, we divide both the equations to find the derivative of ${x^3}$ with respect to ${x^2}$.
So, we get,
$ \Rightarrow \dfrac{{\dfrac{{d\left( {{x^3}} \right)}}{{dx}}}}{{\dfrac{{d\left( {{x^2}} \right)}}{{dx}}}} = \dfrac{{3{x^2}}}{{2x}}$
Cancelling the common factors in numerator and denominator, we get,
$ \Rightarrow \dfrac{{d\left( {{x^3}} \right)}}{{d\left( {{x^2}} \right)}} = \dfrac{{3x}}{2}$
So, the derivative of ${x^3}$ with respect to ${x^2}$ is $\dfrac{{3x}}{2}$.
Therefore, option (A) is the correct answer to the problem.
Note: Derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite function and examining the behaviour of function layer by layer. We must take care of the calculations while solving such questions.
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