Question

Find :$\mathsf {\frac{dy}{dx} (3x^{2} - 9x + 5)^{9}}$
Require thorough explanation. Thanks in advance!

Answer 1

$\sf{\qquad\qquad\huge\underline{{\sf Answer}}}$

Let's solve ~

$\qquad \sf  \dashrightarrow \:\sf \: \dfrac{d}{dx} (3 {x}^{2} - 9x + 5) {}^{9}$

we know :

$\qquad \sf  \dashrightarrow \:\sf \: \dfrac{d}{dx} ({x}^{n}) = nx {}^{n - 1}$

So, by using the folowing property with chain rule ~

$\qquad \sf  \dashrightarrow \:\mathsf { 9(3x^{2} - 9x + 5)^{8}} \sdot(6x - 9)$

Answer 2

Answer:

$\dfrac{dy}{dx}=9(6x-9)(3x^2-9x+5)^8$

Step-by-step explanation:

Given equation:

$y=(3x^2-9x+5)^9$

To differentiate the given equation, use the chain rule:

Chain Rule

$\textsf{If }\: y=f(u) \:\textsf{ and } \: u=g(x) \textsf{ then}:$

$\dfrac{dy}{dx}=\dfrac{dy}{du} \times \dfrac{du}{dx}$

$\textsf{Find } \:\dfrac{dy}{dx}\: \textsf{ if } \:y=(3x^2-9x+5)^9$

$\textsf{Let }\:y=u^9 \: \textsf{ where }\: u=3x^2-9x+5$

Differentiate the two parts separately:

$\implies \dfrac{dy}{du}=9u^8 \quad \textsf{and} \quad \dfrac{du}{dx}=6x-9$

Put everything back into the chain rule formula:

$\begin{aligned}\dfrac{dy}{dx} & =\dfrac{dy}{du} \times \dfrac{du}{dx}\\\\\implies \dfrac{dy}{dx} & = (6x-9) \times 9u^8\\\\& = (6x-9) \times 9(3x^2-9x+5)^8\\\\& = 9(6x-9)(3x^2-9x+5)^8\end{aligned}$