What is the derivative of y=sqrt((a^4)/(a^3-8x)) ?

The derivative of y=sqrt((a^4)/(a^3-8x)) is (4a^4)/(sqrt(a^4)(-8x+a^3)^{3/2)}

What is the first derivative of y=sqrt((a^4)/(a^3-8x)) ?

The first derivative of y=sqrt((a^4)/(a^3-8x)) is (4a^4)/(sqrt(a^4)(-8x+a^3)^{3/2)}

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derivative of

y = \frac{a ^{4}}{a ^{3} - 8 x}

\frac{4 a ^{4}}{a ^{4} ( - 8 x + a ^{3} ) ^{\frac{3}{2}}}

Solution steps

\frac{d}{d x} (\frac{a ^{4}}{a ^{3} - 8 x})

Apply the chain rule:

\frac{1}{2 \frac{a ^{4}}{a ^{3} - 8 x}} \frac{d}{d x} (\frac{a ^{4}}{a ^{3} - 8 x})

= \frac{1}{2 \frac{a ^{4}}{a ^{3} - 8 x}} \frac{d}{d x} (\frac{a ^{4}}{a ^{3} - 8 x})

\frac{d}{d x} (\frac{a ^{4}}{a ^{3} - 8 x}) = \frac{8 a ^{4}}{( a ^{3} - 8 x ) ^{2}}

= \frac{1}{2 \frac{a ^{4}}{a ^{3} - 8 x}} \cdot \frac{8 a ^{4}}{( a ^{3} - 8 x ) ^{2}}

Simplify

\frac{1}{2 \frac{a ^{4}}{a ^{3} - 8 x}} \cdot \frac{8 a ^{4}}{( a ^{3} - 8 x ) ^{2}} : \frac{4 a ^{4}}{a ^{4} ( - 8 x + a ^{3} ) ^{\frac{3}{2}}}

= \frac{4 a ^{4}}{a ^{4} ( - 8 x + a ^{3} ) ^{\frac{3}{2}}}

\frac{\partial}{\partial z} (z y^{2} (t + 1))

\int \frac{sec ( y )}{sin ( y )} d y

\frac{\partial}{\partial x} (\frac{1}{x ^{2}})

\frac{2}{( - 2 s + 10 )}

\frac{d}{d x} (f (x) (x))

What is the derivative of y=sqrt((a^4)/(a^3-8x)) ?
The derivative of y=sqrt((a^4)/(a^3-8x)) is (4a^4)/(sqrt(a^4)(-8x+a^3)^{3/2)}
What is the first derivative of y=sqrt((a^4)/(a^3-8x)) ?
The first derivative of y=sqrt((a^4)/(a^3-8x)) is (4a^4)/(sqrt(a^4)(-8x+a^3)^{3/2)}