What is the d/(dy)(2sqrt(1-y^2-(x/4)^2)) ?

The d/(dy)(2sqrt(1-y^2-(x/4)^2)) is -(8y)/(sqrt(16-16y^2-x^2))

What is the first d/(dy)(2sqrt(1-y^2-(x/4)^2)) ?

The first d/(dy)(2sqrt(1-y^2-(x/4)^2)) is -(8y)/(sqrt(16-16y^2-x^2))

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Popular >

\frac{d}{d y} (2 1 - y^{2} - (\frac{x}{4})^{2})

- \frac{8 y}{16 - 16 y ^{2} - x ^{2}}

Solution steps

\frac{d}{d y} (2 1 - y^{2} - (\frac{x}{4})^{2})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d y} (1 - y^{2} - (\frac{x}{4})^{2})

Apply the chain rule:

\frac{1}{2 1 - y ^{2} - ( \frac{x}{4} ) ^{2}} \frac{d}{d y} (1 - y^{2} - (\frac{x}{4})^{2})

= \frac{1}{2 1 - y ^{2} - ( \frac{x}{4} ) ^{2}} \frac{d}{d y} (1 - y^{2} - (\frac{x}{4})^{2})

\frac{d}{d y} (1 - y^{2} - (\frac{x}{4})^{2}) = - 2 y

= 2 \cdot \frac{1}{2 1 - y ^{2} - ( \frac{x}{4} ) ^{2}} (- 2 y)

Simplify

2 \cdot \frac{1}{2 1 - y ^{2} - ( \frac{x}{4} ) ^{2}} (- 2 y) : - \frac{8 y}{16 - 16 y ^{2} - x ^{2}}

= - \frac{8 y}{16 - 16 y ^{2} - x ^{2}}

f (x) = 17.19 8^{2} - x^{2}

\int sec^{3} (π x) d x

\frac{d}{d x} (e^{x e^{y}})

\frac{d}{d x} (- cot (2 x))

\frac{d}{d x} (cos (x^{2}))

What is the d/(dy)(2sqrt(1-y^2-(x/4)^2)) ?
The d/(dy)(2sqrt(1-y^2-(x/4)^2)) is -(8y)/(sqrt(16-16y^2-x^2))
What is the first d/(dy)(2sqrt(1-y^2-(x/4)^2)) ?
The first d/(dy)(2sqrt(1-y^2-(x/4)^2)) is -(8y)/(sqrt(16-16y^2-x^2))