Implicit Differentiation Made Easy: Expert Advice

Recent questions in Implicit Differentiation

lilmoore11p8 2022-07-15

Use implicit differentiation to find the derivative of $y = \arccos (\sqrt{x})$ as a function of x and say where this derivative is defined.
Don't really get the concept of implicit differentiation.

suchonosdy 2022-07-14

I am trying to implicitly differentiate
$\sin (x / y) = 1 / 2$
The solution manual says
Step 1.
$\cos (x / y) \cdot \frac{y - x \frac{d y}{d x}}{y^{2}} = 0$
But I don't understand how they arrive at this next part:
Step 2.
$y - x \frac{d y}{d x} = 0$
Is $\cos (x / y) = y^{2}$ ?

Implicit differentiation is a method used to find the derivative of an equation in which the equation is not written in the form "y = something". It involves differentiating each side of the equation with respect to the same variable, typically "x", and is useful for finding derivatives of equations with multiple variables and higher-order derivatives. Examples include finding derivatives of equations involving circles, lines, and parabolas. By practicing implicit differentiation, students can learn to solve complex equations and understand the relationship between a function and its derivative.

Calculus 2

Implicit Differentiation Explained: In-Depth Examples, Equations, and Expert Insights

Use implicit differentiation to find the derivative of y = arccos ⁡ ( x ) as a function of x and say where this derivative is defined.Don't really get the concept of implicit differentiation.

I am trying to implicitly differentiate sin ⁡ ( x / y ) = 1 / 2The solution manual saysStep 1. cos ⁡ ( x / y ) ⋅ y − x d y d x y 2 = 0But I don't understand how they arrive at this next part:Step 2. y − x d y d x = 0Is cos ⁡ ( x / y ) = y 2 ?

Use implicit differentiation to find the derivative of $y = \arccos (\sqrt{x})$ as a function of x and say where this derivative is defined.
Don't really get the concept of implicit differentiation.

I am trying to implicitly differentiate
$\sin (x / y) = 1 / 2$
The solution manual says
Step 1.
$\cos (x / y) \cdot \frac{y - x \frac{d y}{d x}}{y^{2}} = 0$
But I don't understand how they arrive at this next part:
Step 2.
$y - x \frac{d y}{d x} = 0$
Is $\cos (x / y) = y^{2}$ ?