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}

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

x \to 2 - lim (x + 2)

\frac{d}{d x} (ln (x + 1) - \frac{1}{x + 1})

\int \frac{1}{5 + x} d x

sin (2 x) d x + cos (5 y) d y = 0, y (\frac{π}{2}) = \frac{π}{5}

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