derivative of ln|sqrt(e^{1+x^2)+3x^4}+cot^2(pix|)

Lösung

\frac{d}{d x} (ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

\frac{x ( 6 x ^{2} + e ^{x^{2} + 1} ) - 2 π csc ^{2} ( π x ) cot ( π x ) 3 x ^{4} + e ^{x^{2} + 1}}{( cot ^{2} ( π x ) + 3 x ^{4} + e ^{x^{2} + 1} ) 3 x ^{4} + e ^{x^{2} + 1}}

Schritte zur Lösung

\frac{d}{d x} (ln e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

Wende die Kettenregel an:

\frac{1}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x )} \frac{d}{d x} (e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

= \frac{1}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x )} \frac{d}{d x} (e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x))

\frac{d}{d x} (e^{1 + x^{2}} + 3 x^{4} + cot^{2} (π x)) = \frac{( x ( 6 x ^{2} + e ^{x^{2} + 1} ) - 2 π csc ^{2} ( π x ) cot ( π x ) e ^{x^{2} + 1} + 3 x ^{4} ) ( e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x ) )}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x ) e ^{x^{2} + 1} + 3 x ^{4}}

= \frac{1}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x )} \cdot \frac{( x ( 6 x ^{2} + e ^{x^{2} + 1} ) - 2 π csc ^{2} ( π x ) cot ( π x ) e ^{x^{2} + 1} + 3 x ^{4} ) ( e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x ) )}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x ) e ^{x^{2} + 1} + 3 x ^{4}}

Vereinfache

\frac{1}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x )} \cdot \frac{( x ( 6 x ^{2} + e ^{x^{2} + 1} ) - 2 π csc ^{2} ( π x ) cot ( π x ) e ^{x^{2} + 1} + 3 x ^{4} ) ( e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x ) )}{e ^{1 + x^{2}} + 3 x ^{4} + cot ^{2} ( π x ) e ^{x^{2} + 1} + 3 x ^{4}} : \frac{x ( 6 x ^{2} + e ^{x^{2} + 1} ) - 2 π csc ^{2} ( π x ) cot ( π x ) 3 x ^{4} + e ^{x^{2} + 1}}{( cot ^{2} ( π x ) + 3 x ^{4} + e ^{x^{2} + 1} ) 3 x ^{4} + e ^{x^{2} + 1}}

= \frac{x ( 6 x ^{2} + e ^{x^{2} + 1} ) - 2 π csc ^{2} ( π x ) cot ( π x ) 3 x ^{4} + e ^{x^{2} + 1}}{( cot ^{2} ( π x ) + 3 x ^{4} + e ^{x^{2} + 1} ) 3 x ^{4} + e ^{x^{2} + 1}}

(\frac{1}{1 + e ^{- x}})^{^{'}}

p a r i t y \frac{tan ( x )}{x}

\int x - x + 2 d x

\int \frac{x}{( 4 x ^{2} + 3 ) ^{6}} d x

t an g e n t \frac{3}{2 3 x + 7}, at x = 6