What is the derivative of e^{sqrt(2x+y^2)} ?

The derivative of e^{sqrt(2x+y^2)} is (e^{sqrt(2x+y^2)}(1+y\frac{dy)/(dx))}{sqrt(2x+y^2)}

What is the first derivative of e^{sqrt(2x+y^2)} ?

The first derivative of e^{sqrt(2x+y^2)} is (e^{sqrt(2x+y^2)}(1+y\frac{dy)/(dx))}{sqrt(2x+y^2)}

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Popular >

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

\frac{e ^{2 x + y^{2}} ( 1 + y \frac{d y}{d x} )}{2 x + y ^{2}}

Solution steps

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

Apply the chain rule:

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

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

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

= e^{2 x + y^{2}} \frac{1 + y \frac{d y}{d x}}{2 x + y ^{2}}

Multiply fractions:

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

= \frac{e ^{2 x + y^{2}} ( 1 + y \frac{d y}{d x} )}{2 x + y ^{2}}

50 \cdot 1 0^{- 3} \cdot (\frac{7.85 \cdot 1 0 ^{- 7}}{2})

(x - 1)^{2} \cdot e^{- 2 x^{^{'}}}

(3 \cdot 1 0^{6}) \cdot (2.7 \cdot 1 0^{4})

1.74 \cdot 1 0^{- 61}

What is the derivative of e^{sqrt(2x+y^2)} ?
The derivative of e^{sqrt(2x+y^2)} is (e^{sqrt(2x+y^2)}(1+y\frac{dy)/(dx))}{sqrt(2x+y^2)}
What is the first derivative of e^{sqrt(2x+y^2)} ?
The first derivative of e^{sqrt(2x+y^2)} is (e^{sqrt(2x+y^2)}(1+y\frac{dy)/(dx))}{sqrt(2x+y^2)}