What is the derivative of ln((1+sqrt(x)/(x^4))) ?

The derivative of ln((1+sqrt(x)/(x^4))) is (x^{7/2}-8x^3(sqrt(x)+1))/(2x^4(sqrt(x)+1))

What is the first derivative of ln((1+sqrt(x)/(x^4))) ?

The first derivative of ln((1+sqrt(x)/(x^4))) is (x^{7/2}-8x^3(sqrt(x)+1))/(2x^4(sqrt(x)+1))

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

\frac{x ^{\frac{7}{2}} - 8 x ^{3} ( x + 1 )}{2 x ^{4} ( x + 1 )}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{1 + x}{x ^{4}}) = \frac{x ^{\frac{7}{2}} - 8 x ^{3} ( 1 + x )}{2 x ^{8}}

= \frac{1}{\frac{1 + x}{x ^{4}}} \cdot \frac{x ^{\frac{7}{2}} - 8 x ^{3} ( 1 + x )}{2 x ^{8}}

Simplify

\frac{1}{\frac{1 + x}{x ^{4}}} \cdot \frac{x ^{\frac{7}{2}} - 8 x ^{3} ( 1 + x )}{2 x ^{8}} : \frac{x ^{\frac{7}{2}} - 8 x ^{3} ( x + 1 )}{2 x ^{4} ( x + 1 )}

= \frac{x ^{\frac{7}{2}} - 8 x ^{3} ( x + 1 )}{2 x ^{4} ( x + 1 )}

n = 6 \sum \infty \frac{3 ^{n}}{1 2 ^{n}}

\int_{0}^{15} t + 1 d t

\frac{d}{d x} ((1 + x^{3} Y^{2} Z)^{2})

\frac{d}{d x} (4^{x} ln (4))

y^{^{'}} (y - x^{2}) = - 2 y

What is the derivative of ln((1+sqrt(x)/(x^4))) ?
The derivative of ln((1+sqrt(x)/(x^4))) is (x^{7/2}-8x^3(sqrt(x)+1))/(2x^4(sqrt(x)+1))
What is the first derivative of ln((1+sqrt(x)/(x^4))) ?
The first derivative of ln((1+sqrt(x)/(x^4))) is (x^{7/2}-8x^3(sqrt(x)+1))/(2x^4(sqrt(x)+1))