What is the derivative of y=16\sqrt[4]{4x^4+4} ?

The derivative of y=16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}

What is the first derivative of y=16\sqrt[4]{4x^4+4} ?

The first derivative of y=16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}

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derivative of

\frac{64 x ^{3}}{( 4 x ^{4} + 4 ) ^{\frac{3}{4}}}

Solution steps

Take the constant out:

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

Apply the chain rule:

\frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \frac{d}{d x} (4 x^{4} + 4)

= \frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \frac{d}{d x} (4 x^{4} + 4)

\frac{d}{d x} (4 x^{4} + 4) = 16 x^{3}

= 16 \cdot \frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \cdot 16 x^{3}

Simplify

16 \cdot \frac{1}{4 ( 4 x ^{4} + 4 ) ^{\frac{3}{4}}} \cdot 16 x^{3} : \frac{64 x ^{3}}{( 4 x ^{4} + 4 ) ^{\frac{3}{4}}}

= \frac{64 x ^{3}}{( 4 x ^{4} + 4 ) ^{\frac{3}{4}}}

x \to 2 lim (5 x^{3} - 7^{2} + 6)

\frac{e ^{x}}{x}

\frac{\partial}{\partial x} (y^{3} - ln (x))

\int 4 x \cdot e^{x} d x

\frac{\partial}{\partial x} (- \frac{9 x}{y})

What is the derivative of y=16\sqrt[4]{4x^4+4} ?
The derivative of y=16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}
What is the first derivative of y=16\sqrt[4]{4x^4+4} ?
The first derivative of y=16\sqrt[4]{4x^4+4} is (64x^3)/((4x^4+4)^{3/4)}