• Algebra
• Trigonometry
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• Calculus

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 $t \,=\, 2x+3$

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