Please help :) Due asap and thx!

If -y-2x^3=Y^2 then find D^2y/dx^2 at the point (-1,-2) in simplest form

algol13

Answer:

$\frac{d^2y}{dx^2} = \frac{-4}{3}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Algebra I

Factoring

Calculus

Implicit Differentiation

The derivative of a constant is equal to 0

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Product Rule: $\frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)$

Chain Rule: $\frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Quotient Rule: $\frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

-y - 2x³ = y²

Rate of change of tangent line at point (-1, -2)

Step 2: Differentiate Pt. 1

Find 1st Derivative

Implicit Differentiation [Basic Power Rule]: $-y'-6x^2=2yy'$
[Algebra] Isolate y' terms: $-6x^2=2yy'+y'$
[Algebra] Factor y': $-6x^2=y'(2y+1)$
[Algebra] Isolate y': $\frac{-6x^2}{(2y+1)}=y'$
[Algebra] Rewrite: $y' = \frac{-6x^2}{(2y+1)}$

Step 3: Differentiate Pt. 2

Find 2nd Derivative

Differentiate [Quotient Rule/Basic Power Rule]: $y'' = \frac{-12x(2y+1)+6x^2(2y')}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x+12x^2y'}{(2y+1)^2}$
[Derivative] Back-Substitute y': $y'' = \frac{-24xy-12x+12x^2(\frac{-6x^2}{2y+1} )}{(2y+1)^2}$
[Derivative] Simplify: $y'' = \frac{-24xy-12x-\frac{72x^4}{2y+1} }{(2y+1)^2}$

Step 4: Find Slope at Given Point

[Algebra] Substitute in x and y: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(-1)^4}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-24(-1)(-2)-12(-1)-\frac{72(1)}{2(-2)+1} }{(2(-2)+1)^2}$
[Pre-Algebra] Multiply: $y''(-1,-2) = \frac{-48+12-\frac{72}{-4+1} }{(-4+1)^2}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{(-3)^2}$
[Pre-Algebra] Exponents: $y''(-1,-2) = \frac{-36-\frac{72}{-3} }{9}$
[Pre-Algebra] Divide: $y''(-1,-2) = \frac{-36+24 }{9}$
[Pre-Algebra] Add: $y''(-1,-2) = \frac{-12}{9}$
[Pre-Algebra] Simplify: $y''(-1,-2) = \frac{-4}{3}$

6 0

2 years ago

What is the solution to -13×=78

Pavlova-9 [17]

-13x=78
-13x/-13=78/-13
Answers: X=-6

8 0

3 years ago

Read 2 more answers

Write the standard form of the equation of a circle with radius 10 and center (9,3)

gladu [14]

Answer:

$(x-9)^{2}+(y-3)^{2} =100$

Step-by-step explanation:

The standard form formula for a circle equation is $(x-h)^{2}+(y-k)^{2} =r^{2}$ , where (h,k) represents the center of the circle, and r represents the radius.

r=10

(h,k)= (9,3)

$(x-h)^{2}+(y-k)^{2} =r^{2}$

$(x-9)^{2}+(y-3)^{2} =10^{2}\\(x-9)^{2}+(y-3)^{2} =100$

8 0

2 years ago

The diameter of a cylinder is 8cm and the height is 2. What is the Volume? (NOTE: V=r2h)

Afina-wow [57]

Step-by-step explanation:

diameter= 8

radius=diameter/2

=8/2

=4

height=2

Volume =r2h

=4*2*2

=16

therefore the volume of the cylinder is 16 cm^3

hope it helps u

pls mark it as the brainliest

3 0

2 years ago

(2-3) exponent 4 (4)-9 divided by 3

Ilya [14]

Answer:

The correct answer is 1.

Step-by-step explanation:

Hope this helped Mark BRAINLIEST!!!

3 0

2 years ago