What is (\partial)/(\partial x)(2y^{4x}) ?

The answer to (\partial)/(\partial x)(2y^{4x}) is 8y^{4x}ln(y)

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(\partial)/(\partial x)(2y^{4x})

\frac{\partial}{\partial x} (2 y^{4 x})

8 y^{4 x} ln (y)

Solution steps

\frac{\partial}{\partial x} (2 y^{4 x})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply exponent rule:

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

y^{4 x} = e^{4 x l n (y)}

= 2 \frac{\partial}{\partial x} (e^{4 x l n (y)})

Apply the chain rule:

e^{4 x l n (y)} \frac{\partial}{\partial x} (4 x ln (y))

= e^{4 x l n (y)} \frac{\partial}{\partial x} (4 x ln (y))

\frac{\partial}{\partial x} (4 x ln (y)) = 4 ln (y)

= 2 e^{4 x l n (y)} \cdot 4 ln (y)

Simplify

2 e^{4 x l n (y)} \cdot 4 ln (y) : 8 y^{4 x} ln (y)

= 8 y^{4 x} ln (y)

f (x) = sin (x) + 1, at x = - \frac{π}{2}

f (x) = x^{2} - 4, x = - 4, x = 3

\int \frac{1}{x ^{2} a ^{2} - x ^{2}} d x

x \to - 3 lim (∣ x ∣ - 2)

e^{t}

What is (\partial)/(\partial x)(2y^{4x}) ?
The answer to (\partial)/(\partial x)(2y^{4x}) is 8y^{4x}ln(y)