What is the derivative of sqrt(((x+1/(x+6))^8)) ?

The derivative of sqrt(((x+1/(x+6))^8)) is (20(x+1)^3)/((x+6)^5)

What is the first derivative of sqrt(((x+1/(x+6))^8)) ?

The first derivative of sqrt(((x+1/(x+6))^8)) is (20(x+1)^3)/((x+6)^5)

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\frac{d}{d x} (\frac{x + 1}{x + 6})^{8}

\frac{20 ( x + 1 ) ^{3}}{( x + 6 ) ^{5}}

Solution steps

\frac{d}{d x} (\frac{x + 1}{x + 6})^{8}

Apply the chain rule:

\frac{1}{2 ( \frac{x + 1}{x + 6} ) ^{8}} \frac{d}{d x} ((\frac{x + 1}{x + 6})^{8})

= \frac{1}{2 ( \frac{x + 1}{x + 6} ) ^{8}} \frac{d}{d x} ((\frac{x + 1}{x + 6})^{8})

\frac{d}{d x} ((\frac{x + 1}{x + 6})^{8}) = \frac{40 ( x + 1 ) ^{7}}{( x + 6 ) ^{9}}

= \frac{1}{2 ( \frac{x + 1}{x + 6} ) ^{8}} \cdot \frac{40 ( x + 1 ) ^{7}}{( x + 6 ) ^{9}}

Simplify

\frac{1}{2 ( \frac{x + 1}{x + 6} ) ^{8}} \cdot \frac{40 ( x + 1 ) ^{7}}{( x + 6 ) ^{9}} : \frac{20 ( x + 1 ) ^{3}}{( x + 6 ) ^{5}}

= \frac{20 ( x + 1 ) ^{3}}{( x + 6 ) ^{5}}

y^{^{'}} = \frac{1}{( t - 1 ) ( y + 1 )}

f (x) = x^{3} (x - 6)^{2}

f (x) = x - 1 + x + 1

y = (\frac{( x + 2 )}{( x - 2 )})^{2}, at x = 3

\int_{0}^{π} 14 sin^{4} (x) d x

What is the derivative of sqrt(((x+1/(x+6))^8)) ?
The derivative of sqrt(((x+1/(x+6))^8)) is (20(x+1)^3)/((x+6)^5)
What is the first derivative of sqrt(((x+1/(x+6))^8)) ?
The first derivative of sqrt(((x+1/(x+6))^8)) is (20(x+1)^3)/((x+6)^5)