What is the derivative of y= 1/(\sqrt[4]{3x^2+7)} ?

The derivative of y= 1/(\sqrt[4]{3x^2+7)} is -(3x)/(2(3x^2+7)^{5/4)}

What is the first derivative of y= 1/(\sqrt[4]{3x^2+7)} ?

The first derivative of y= 1/(\sqrt[4]{3x^2+7)} is -(3x)/(2(3x^2+7)^{5/4)}

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derivative of

- \frac{3 x}{2 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}}

Solution steps

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} ((3 x^{2} + 7)^{- \frac{1}{4}})

Apply the chain rule:

- \frac{1}{4 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}} \frac{d}{d x} (3 x^{2} + 7)

= - \frac{1}{4 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}} \frac{d}{d x} (3 x^{2} + 7)

\frac{d}{d x} (3 x^{2} + 7) = 6 x

= - \frac{1}{4 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}} \cdot 6 x

Simplify

- \frac{1}{4 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}} \cdot 6 x : - \frac{3 x}{2 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}}

= - \frac{3 x}{2 ( 3 x ^{2} + 7 ) ^{\frac{5}{4}}}

\frac{d}{d x} (- f (x) sin (x) + g (x) cos (x))

\int \frac{x + 3}{5 - 4 x - x ^{2}} d x

\frac{d ^{2}}{d x ^{2}} (\frac{( x + b ) ( x - a )}{( x - b )})

(cos (x) sin (x) - x y^{2} 2) d x + y (1 - x^{2}) d y = 0

\frac{3 x}{2 - cot ( x )}

What is the derivative of y= 1/(\sqrt[4]{3x^2+7)} ?
The derivative of y= 1/(\sqrt[4]{3x^2+7)} is -(3x)/(2(3x^2+7)^{5/4)}
What is the first derivative of y= 1/(\sqrt[4]{3x^2+7)} ?
The first derivative of y= 1/(\sqrt[4]{3x^2+7)} is -(3x)/(2(3x^2+7)^{5/4)}