What is (d^4)/(dx^4)(sqrt(1+x)) ?

The answer to (d^4)/(dx^4)(sqrt(1+x)) is -(15)/(16(1+x)^{7/2)}

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(d^4)/(dx^4)(sqrt(1+x))

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

- \frac{15}{16 ( 1 + x ) ^{\frac{7}{2}}}

Solution steps

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

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

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

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

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

\frac{d}{d x} (- \frac{1}{4 ( 1 + x ) ^{\frac{3}{2}}}) = \frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}}

= \frac{d}{d x} (\frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}})

\frac{d}{d x} (\frac{3}{8 ( 1 + x ) ^{\frac{5}{2}}}) = - \frac{15}{16 ( 1 + x ) ^{\frac{7}{2}}}

= - \frac{15}{16 ( 1 + x ) ^{\frac{7}{2}}}

1 0^{5 x^{2} - 3 x}

\int_{1}^{\infty} e^{- x^{2}} d x

y = x^{2}, x = y + 2, - 1, 2

f (x) = \frac{1 + x}{e ^{x}}

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

What is (d^4)/(dx^4)(sqrt(1+x)) ?
The answer to (d^4)/(dx^4)(sqrt(1+x)) is -(15)/(16(1+x)^{7/2)}