Mathematics, 31.10.2020 14:00 kyliegriffis

Second derivative of x^½ + y^½ = a^½

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Answer from: madeleine1102

$\displaystyle y'' = \frac{\sqrt{x} + \sqrt{y}}{2x^\big{\frac{3}{2}}}$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

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

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

Implicit Differentiation

Step-by-step explanation:

*Note:

Allow a to be defined as an arbitrary constant.

Step 1: Define

$\displaystyle x^\big{\frac{1}{2}} + y^\big{\frac{1}{2}} = a^\big{\frac{1}{2}}$

Step 2: Differentiate

Basic Power Rule [Addition/Subtraction, Chain Rule]: $\displaystyle \frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0$ Isolate y' term: $\displaystyle \frac{y'}{2\sqrt{y}} = -\frac{1}{2\sqrt{x}}$ Isolate y': $\displaystyle y' = -\frac{\sqrt{y}}{\sqrt{x}}$ Derivative Rule [Quotient Rule]: $\displaystyle y'' = -\frac{(\sqrt{y})'(\sqrt{x}) - \sqrt{y}(\sqrt{x})'}{(\sqrt{x})^2}$ Basic Power Rule [Derivative Rule - Chain Rule]: $\displaystyle y'' = -\frac{\frac{y'\sqrt{x}}{2\sqrt{y}} - \frac{\sqrt{y}}{2\sqrt{x}}}{(\sqrt{x})^2}$ Simplify: $\displaystyle y'' = -\frac{\frac{y'\sqrt{x}}{2\sqrt{y}} - \frac{\sqrt{y}}{2\sqrt{x}}}{x}$ Rewrite: $\displaystyle y'' = \frac{-(y'x - y)}{2x^\big{\frac{3}{2}}\sqrt{y}}$ Substitute in y': $\displaystyle y'' = \frac{-(-\frac{\sqrt{y}}{\sqrt{x}}x - y)}{2x^\big{\frac{3}{2}}\sqrt{y}}$ Simplify: $\displaystyle y'' = \frac{\sqrt{x} + \sqrt{y}}{2x^\big{\frac{3}{2}}}$

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Differentiation

Answer from: Quest

δabc ≅ δdcb by sas because

db ≅ ac . . given (side)

bc ≅ bc . . reflexive property (side)

∠dbc ≅ ∠ acb . . alternate interior angles where transversal bc crosses parallel lines bd and ac (included angle)

Answer from: Quest

3 ft is your answer

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