What is (\partial)/(\partial x)(2^{x^2-y^5}) ?

The answer to (\partial)/(\partial x)(2^{x^2-y^5}) is ln(2)*2^{x^2-y^5+1}x

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(\partial)/(\partial x)(2^{x^2-y^5})

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

ln (2) \cdot 2^{x^{2} - y^{5} + 1} x

Solution steps

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

Treat

y

as a constant

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{x^{2} - y^{5}} = e^{(x^{2} - y^{5}) l n (2)}

= \frac{\partial}{\partial x} (e^{(x^{2} - y^{5}) l n (2)})

Apply the chain rule:

e^{(x^{2} - y^{5}) l n (2)} \frac{\partial}{\partial x} ((x^{2} - y^{5}) ln (2))

= e^{(x^{2} - y^{5}) l n (2)} \frac{\partial}{\partial x} ((x^{2} - y^{5}) ln (2))

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

= e^{(x^{2} - y^{5}) l n (2)} \cdot 2 ln (2) x

Simplify

e^{(x^{2} - y^{5}) l n (2)} \cdot 2 ln (2) x : ln (2) \cdot 2^{x^{2} - y^{5} + 1} x

= ln (2) \cdot 2^{x^{2} - y^{5} + 1} x

x^{3} - 8 x^{2} + 5 x - 2

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

(7 x^{6} + 2 x^{3})^{4}

x + 8

n = 0 \sum \infty 3^{n} 7^{- n}

What is (\partial)/(\partial x)(2^{x^2-y^5}) ?
The answer to (\partial)/(\partial x)(2^{x^2-y^5}) is ln(2)*2^{x^2-y^5+1}x