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

解

\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}}

解答ステップ

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

連鎖法則を適用:

\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}}

簡素化

\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}}