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Popular >

(\partial)/(\partial y)(1+x^3+y^4)

Solution

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

Solution

4 y^{3}

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{\partial}{\partial y} (1) = 0

\frac{\partial}{\partial y} (x^{3}) = 0

\frac{\partial}{\partial y} (y^{4}) = 4 y^{3}

= 0 + 0 + 4 y^{3}

Simplify

= 4 y^{3}

Popular Examples

(dx)/(dt)=0.25-1/20 x,x(0)=0.5

\frac{d x}{d t} = 0.25 - \frac{1}{20} x, x (0) = 0.5

y^{''}+25y=3sec^3(5)

y^{^{''}} + 25 y = 3 sec^{3} (5)

simplify x/(x^2+y^2)simplify

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

sum from n=1 to infinity of 1/(2^nn^2)

n = 1 \sum \infty \frac{1}{2 ^{n} n ^{2}}

derivative of 2x^{-3}

\frac{d}{d x} (2 x^{- 3})

Frequently Asked Questions (FAQ)

What is (\partial)/(\partial y)(1+x^3+y^4) ?
The answer to (\partial)/(\partial y)(1+x^3+y^4) is 4y^3