What is (\partial)/(\partial x)(ln(x+y^4)) ?

The answer to (\partial)/(\partial x)(ln(x+y^4)) is 1/(x+y^4)

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(\partial)/(\partial x)(ln(x+y^4))

\frac{\partial}{\partial x} (ln (x + y^{4}))

\frac{1}{x + y ^{4}}

Solution steps

\frac{\partial}{\partial x} (ln (x + y^{4}))

Treat

y

as a constant

Apply the chain rule:

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

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

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

= \frac{1}{x + y ^{4}} \cdot 1

Simplify

= \frac{1}{x + y ^{4}}

x \to \infty lim (- \frac{x}{2})

x \to \frac{π}{2} lim (5 sin (x))

f (x) = - 9 x^{2}, at x = 2

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

f (t) = 2 - \frac{4}{t}

What is (\partial)/(\partial x)(ln(x+y^4)) ?
The answer to (\partial)/(\partial x)(ln(x+y^4)) is 1/(x+y^4)