What is the derivative of sqrt(x+y^4) ?

The derivative of sqrt(x+y^4) is (1+4y^3\frac{dy)/(dx)}{2sqrt(x+y^4)}

What is the first derivative of sqrt(x+y^4) ?

The first derivative of sqrt(x+y^4) is (1+4y^3\frac{dy)/(dx)}{2sqrt(x+y^4)}

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Popular >

\frac{d}{d x} (x + y^{4})

\frac{1 + 4 y ^{3} \frac{d y}{d x}}{2 x + y ^{4}}

Solution steps

\frac{d}{d x} (x + y^{4})

Apply the chain rule:

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

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

\frac{d}{d x} (x + y^{4}) = 1 + 4 y^{3} \frac{d y}{d x}

= \frac{1}{2 x + y ^{4}} (1 + 4 y^{3} \frac{d y}{d x})

Simplify

\frac{1}{2 x + y ^{4}} (1 + 4 y^{3} \frac{d y}{d x}) : \frac{1 + 4 y ^{3} \frac{d y}{d x}}{2 x + y ^{4}}

= \frac{1 + 4 y ^{3} \frac{d y}{d x}}{2 x + y ^{4}}

\frac{d}{d t} ((w) (t^{2} w^{2}))

(2 x^{3}) (8 x^{4})

e^{- l n^{2} (x)}

0.02 \cdot 1 0^{- 2}

4.5 \cdot 1 0^{- 1}

What is the derivative of sqrt(x+y^4) ?
The derivative of sqrt(x+y^4) is (1+4y^3\frac{dy)/(dx)}{2sqrt(x+y^4)}
What is the first derivative of sqrt(x+y^4) ?
The first derivative of sqrt(x+y^4) is (1+4y^3\frac{dy)/(dx)}{2sqrt(x+y^4)}