What is the derivative of y=\sqrt[3]{1.6x^2+4.9} ?

The derivative of y=\sqrt[3]{1.6x^2+4.9} is (1.06666…x)/((1.6x^2+4.9)^{2/3)}

What is the first derivative of y=\sqrt[3]{1.6x^2+4.9} ?

The first derivative of y=\sqrt[3]{1.6x^2+4.9} is (1.06666…x)/((1.6x^2+4.9)^{2/3)}

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derivative of

y = 3 1.6 x^{2} + 4.9

\frac{1.06666 \dots x}{( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}}

Solution steps

\frac{d}{d x} (3 1.6 x^{2} + 4.9)

Apply the chain rule:

\frac{1}{3 ( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}} \frac{d}{d x} (1.6 x^{2} + 4.9)

= \frac{1}{3 ( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}} \frac{d}{d x} (1.6 x^{2} + 4.9)

\frac{d}{d x} (1.6 x^{2} + 4.9) = 3.2 x

= \frac{1}{3 ( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}} \cdot 3.2 x

Simplify

\frac{1}{3 ( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}} \cdot 3.2 x : \frac{1.06666 \dots x}{( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}}

= \frac{1.06666 \dots x}{( 1.6 x ^{2} + 4.9 ) ^{\frac{2}{3}}}

f (t) = \frac{1}{2} (9 t^{2} + t)^{- 3}

x \to \infty lim (2 1^{x})

\frac{\partial}{\partial y} (cos (12 x) sin (2 y))

\int \frac{9}{x ln ( 3 x )} d x

f (x) = - 247

What is the derivative of y=\sqrt[3]{1.6x^2+4.9} ?
The derivative of y=\sqrt[3]{1.6x^2+4.9} is (1.06666…x)/((1.6x^2+4.9)^{2/3)}
What is the first derivative of y=\sqrt[3]{1.6x^2+4.9} ?
The first derivative of y=\sqrt[3]{1.6x^2+4.9} is (1.06666…x)/((1.6x^2+4.9)^{2/3)}