How do you use implicit differentiation to find #(dy)/(dx)# given #5y^2=2x^3-5y#?

Question

Scarlett Allison · Accepted Answer

# dy/dx=(6x^2)/(2y+1).#

#5y^2=2x^3-5y rArr 5y^2+5y=5(y^2+y)=2x^3.#

#:. d/dx{5(y^2+y)}=d/dx{2x^3}.#

#:. 5d/dx{y^2+y}=2d/dx{x^3}=2{3x^2}=6x^2...(1),# whereas,

using the Chain Rule,

#d/dx{y^2+y}=d/dx{y^2}+d/dx{y},#

#=d/dy{y^2}*d/dx{y}+dy/dx,#

#=2y*dy/dx+dy/dx.#

Utilising this in #(1),# we get,

# 5{2y*dy/dx+dy/dx}=5(2y+1)dy/dx=6x^2, # giving,

# dy/dx=(6x^2)/{5(2y+1)}#.

Elijah Abram · Answer

undefined To use implicit differentiation to find $ \frac{dy}{dx} $ given $ 5y^2 = 2x^3 - 5y $, follow these steps:

1. Differentiate both sides of the equation with respect to $ x $.
2. Use the chain rule where necessary.
3. Solve the resulting equation for $ \frac{dy}{dx} $.

Differentiating both sides with respect to $ x $ gives:

$$ \frac{d}{dx}(5y^2) = \frac{d}{dx}(2x^3 - 5y) $$

Using the chain rule, differentiate each term:

$$ \frac{d}{dx}(5y^2) = 10y \cdot \frac{dy}{dx} $$

$$ \frac{d}{dx}(2x^3 - 5y) = 6x^2 - 5 \frac{dy}{dx} $$

Rearrange terms to solve for $ \frac{dy}{dx} $:

$$ 10y \cdot \frac{dy}{dx} = 6x^2 - 5 \frac{dy}{dx} $$

$$ 10y \cdot \frac{dy}{dx} + 5 \frac{dy}{dx} = 6x^2 $$

$$ (10y + 5) \frac{dy}{dx} = 6x^2 $$

$$ \frac{dy}{dx} = \frac{6x^2}{10y + 5} $$