What is (\partial)/(\partial x)(yln(x^4+y^3+2)-7) ?

The answer to (\partial)/(\partial x)(yln(x^4+y^3+2)-7) is (4yx^3)/(y^3+x^4+2)

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(\partial)/(\partial x)(yln(x^4+y^3+2)-7)

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2) - 7)

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

Solution steps

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2) - 7)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2)) - \frac{\partial}{\partial x} (7)

\frac{\partial}{\partial x} (y ln (x^{4} + y^{3} + 2)) = \frac{4 y x ^{3}}{x ^{4} + y ^{3} + 2}

\frac{\partial}{\partial x} (7) = 0

= \frac{4 y x ^{3}}{x ^{4} + y ^{3} + 2} - 0

Simplify

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

(\frac{4}{3}) p i x^{3}

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

f (x) = \frac{( x - 1 ) ^{2}}{( x + 2 )}

x sin (y) + y sin (x) = 6

y^{^{'}} + 5 x e^{y} = 0

What is (\partial)/(\partial x)(yln(x^4+y^3+2)-7) ?
The answer to (\partial)/(\partial x)(yln(x^4+y^3+2)-7) is (4yx^3)/(y^3+x^4+2)