What is (\partial)/(\partial y)(x^5+2^y+x^y) ?

The answer to (\partial)/(\partial y)(x^5+2^y+x^y) is x^yln(x)+ln(2)*2^y

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(\partial)/(\partial y)(x^5+2^y+x^y)

\frac{\partial}{\partial y} (x^{5} + 2^{y} + x^{y})

x^{y} ln (x) + ln (2) \cdot 2^{y}

Solution steps

\frac{\partial}{\partial y} (x^{5} + 2^{y} + x^{y})

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial y} (x^{5}) + \frac{\partial}{\partial y} (2^{y}) + \frac{\partial}{\partial y} (x^{y})

\frac{\partial}{\partial y} (x^{5}) = 0

\frac{\partial}{\partial y} (2^{y}) = 2^{y} ln (2)

\frac{\partial}{\partial y} (x^{y}) = x^{y} ln (x)

= 0 + 2^{y} ln (2) + x^{y} ln (x)

Simplify

= x^{y} ln (x) + ln (2) \cdot 2^{y}

x \to 2 - lim (x + 2)

y^{^{''''}} + 20 y^{^{''}} + 100 y = 0

y^{^{'}} = e^{5 x + 3 y}

f (t) = cos^{3} (t)

y^{^{''}} - y^{^{'}} - 2 y = e^{3 x} sin (2 x)

What is (\partial)/(\partial y)(x^5+2^y+x^y) ?
The answer to (\partial)/(\partial y)(x^5+2^y+x^y) is x^yln(x)+ln(2)*2^y