What is the derivative of y^2sqrt(9+x^2) ?

The derivative of y^2sqrt(9+x^2) is 2y(dy)/(dx)sqrt(9+x^2)+(xy^2)/(sqrt(9+x^2))

What is the first derivative of y^2sqrt(9+x^2) ?

The first derivative of y^2sqrt(9+x^2) is 2y(dy)/(dx)sqrt(9+x^2)+(xy^2)/(sqrt(9+x^2))

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\frac{d}{d x} (y^{2} 9 + x^{2})

2 y \frac{d y}{d x} 9 + x^{2} + \frac{x y ^{2}}{9 + x ^{2}}

Solution steps

\frac{d}{d x} (y^{2} 9 + x^{2})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = y^{2}, g = 9 + x^{2}

= \frac{d}{d x} (y^{2}) 9 + x^{2} + \frac{d}{d x} (9 + x^{2}) y^{2}

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

\frac{d}{d x} (9 + x^{2}) = \frac{x}{9 + x ^{2}}

= 2 y \frac{d y}{d x} 9 + x^{2} + \frac{x}{9 + x ^{2}} y^{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= 2 y \frac{d y}{d x} 9 + x^{2} + \frac{x y ^{2}}{9 + x ^{2}}

2 x^{3} + 3 x^{2} - 36 x + 90

f (x) = x^{4} - 4 x^{3} - 2

\frac{x ^{2} - 3}{x ^{3}}

4 x^{3} - 8

x^{\frac{4}{5}} - 2

What is the derivative of y^2sqrt(9+x^2) ?
The derivative of y^2sqrt(9+x^2) is 2y(dy)/(dx)sqrt(9+x^2)+(xy^2)/(sqrt(9+x^2))
What is the first derivative of y^2sqrt(9+x^2) ?
The first derivative of y^2sqrt(9+x^2) is 2y(dy)/(dx)sqrt(9+x^2)+(xy^2)/(sqrt(9+x^2))