Math Doubts

Find $\dfrac{d}{dx}{\dfrac{9\pi}{16}{(2x+3)}^3}$

It is an algebraic function, expressed in power form by considering $x$ as a variable in the differential calculus.

Eliminate constant from differentiation

$\dfrac{9\pi}{16}$ is a constant and it can be moved out from the differentiation to go ahead in finding derivative of the function.

$=\,\,\, \dfrac{9\pi}{16}\dfrac{d}{dx}{{(2x+3)}^3}$

Simplifying the Power of the function

$2x+3$ is raised to the power of $3$ is a function in power form but there is no differentiation rule to find the derivative of this function.

Actually, a differentiation rule of the derivative of the function $x^n$ with respect to $x$ is known in the differential calculus. But, there is a lot of difference between functions $x^n$ and ${(2x+3)}^3$. If the function ${(2x+3)}^3$ is transformed in the form of $x^n$, then the differentiation formula of $x^n$ can be applied.

Take, $t \,=\, 2x+3$, then differentiate both sides of the equation with respect to $x$.

$\implies \dfrac{dt}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(2x+3)}$

$\implies \dfrac{dt}{dx}$ $\,=\,$ $\dfrac{d}{dx}{(2x)}+\dfrac{d}{dx}{(3)}$

The derivative of a variable with respect to same variable is equal to one and the derivative of a constant is always zero.

$\implies \dfrac{dt}{dx}$ $\,=\,$ $2\dfrac{d}{dx}{x}+0$

$\implies \dfrac{dt}{dx}$ $\,=\,$ $2 \times 1$

$\implies dt$ $\,=\,$ $2dx$

$\,\,\, \therefore \,\,\,\,\,\, dx$ $\,=\,$ $\dfrac{dt}{2}$

Now, transform the derivative of the function by eliminating the $x$ by $t$.

We know that, $t \,=\, 2x+3$ and $dx$ $\,=\,$ $\dfrac{dt}{2}$.

$\implies \dfrac{9\pi}{16}\dfrac{d}{dx}{{(2x+3)}^3}$ $\,=\,$ $\dfrac{9\pi}{16}\dfrac{d}{\dfrac{dt}{2}}{{(t)}^3}$

$\implies \dfrac{9\pi}{16}\dfrac{d}{dx}{{(2x+3)}^3}$ $\,=\,$ $\dfrac{9\pi}{16}\dfrac{2d}{dt}{t^3}$

$\implies \dfrac{9\pi}{16}\dfrac{d}{dx}{{(2x+3)}^3}$ $\,=\,$ $\dfrac{9\pi \times 2}{16}\dfrac{d}{dt}{t}^3$

$\,=\,$ $\require{cancel} \dfrac{9\pi \times \cancel{2}}{\cancel{16}}\dfrac{d}{dt}{t}^3$

$\,=\,$ $\dfrac{9\pi}{8}\dfrac{d}{dt}{t}^3$

Differentiate the function

As per derivative of $x^n$ rule, the differentiation of $t^3$ with respect to $t$ can be performed.

$\,=\,$ $\dfrac{9\pi}{8} \times 3t^{(3-1)}$

$\,=\,$ $\dfrac{9\pi \times 3}{8}t^2$

$\,=\,$ $\dfrac{27\pi}{8}t^2$

Bring back the function to x terms

The solution of the function is in terms of $t$ but the derivative of the function is given in terms of $x$. So, it is very important to convert the function in terms of $x$ in order to finish this differentiation problem.

It is considered that $t \,=\, 2x+3$

$\,\,\, \therefore \,\,\,\,\,\, \dfrac{d}{dx}{\dfrac{9\pi}{16}{(2x+3)}^3}$ $\,=\,$ $\dfrac{27\pi}{8}{(2x+3)}^2$

Math Doubts
Math Doubts is a free math tutor for helping students to learn mathematics online from basics to advanced scientific level for teachers to improve their teaching skill and for researchers to share their research projects. Know more
Follow us on Social Media
Math Problems

Learn how to solve easy to difficult mathematics problems of all topics in various methods with step by step process and also maths questions for practising.

Learn more