Question

$\huge{Question}$​

Answer 1

Answer:

${\boxed{\boxed{\begin{array}{cc}\bf \: \to \:given : \\ \\ \rm \: y = 2 {x}^{3} - 3 {x}^{2} - 36x + 2 \\ \\ \sf \: we \: have \: to \: differentiate \: it \: w.r.t. \: \: \bf \: x \\ \\ \therefore \: \rm \: \frac{dy}{dx} = \frac{d}{dx} (2 {x}^{3} - 3 {x}^{2} - 36x + 2) \\ \\ \pink{{\boxed{\begin{array}{cc}\bf \:we \: know \: that : \\ \bf \: \frac{d}{dx}(u \pm \: v) = \frac{d}{dx} \: u \pm \frac{d}{dx} \: v\end{array}}}} \\ \orange{ \sf \: apply \: this} \\ \\ \rm = \frac{d}{dx}2 {x}^{3} - \frac{d}{dx} 3 {x}^{2} - \frac{d}{dx} 36x + \frac{d}{dx} \: 2 \\ \\ \pink{{\boxed{\begin{array}{cc}\bf \: we \: know \: that : \\ \bf \: \frac{d}{dx} \: [constant(c)] = 0 \\ \\ \bf \: \frac{d}{dx} \: c {x}^{n} = cn {x}^{n - 1} \end{array}}}} \\ \orange{ \sf \: apply \: this} \\ \\ \rm = 2 \times 3 \times {x}^{3 - 1} - 3 \times 2 \times {x}^{2 - 1} - 36 \times {x}^{1 - 1} + 0 \\ \\ \rm = 6 {x}^{2} - 6x - 36 \\ \\ \\ \\ \blue{ \boxed{\therefore \rm \: \frac{dy}{dx} = 6 {x}^{2} - 6x - 36}}\end{array}}}}$

Answer 2

Explanation:

Hey mate

Here is ur answer

y=2x³+3x²-36x+7 : Given

dy/dx = 0 : Given

Find y

dy/dx = 2x³+3x²-36x+7

0 =2*3*x3–1 + 3*2*x2–1 –36*x1–1 +0

0 = 6x2 + 6x - 36

0 = x2 + x -6

0 = x2 + 3x - 2x -6

0= x(x+3) -2(x+3)

0 = (x-2) (x+3)

x-2 = 0 x= 2 ; x+3=0 x= -3

y=2x³+3x²-36x+7

Consider x=2

y= 2(2)3 + 3(2)2 -36*2 +7

y= 16+12–72+7

y= 35–72

y= 37 : Final Answer