(t/(t+1(sqrt(t))))^'

解答

(\frac{t}{t + 1 ( t )})^{^{'}}

\frac{t}{2 ( t + t ) ^{2}}

求解步骤

(\frac{t}{t + 1 \cdot ( t )})^{'}

使用微分除法定则:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{t ^{^{'}} ( t + 1 \cdot t ) - ( t + 1 \cdot t ) ^{^{'}} t}{( t + 1 \cdot t ) ^{2}}

t^{'} = 1

(t + 1 \cdot t)^{'} = 1 + \frac{1}{2 t}

= \frac{1 \cdot ( t + 1 \cdot t ) - ( 1 + \frac{1}{2 t} ) t}{( t + 1 \cdot t ) ^{2}}

化简

\frac{1 \cdot ( t + 1 \cdot t ) - ( 1 + \frac{1}{2 t} ) t}{( t + 1 \cdot t ) ^{2}} : \frac{t}{2 ( t + t ) ^{2}}

= \frac{t}{2 ( t + t ) ^{2}}

y^{^{'}} + \frac{y}{x} = \frac{1}{x ^{2}}

t an g e n t y = 4 sin (π x + y), (1, 0)

n = 0 \sum \infty 6 (- \frac{2}{3})^{n}

y^{^{'}} = \frac{y ^{4} - 3}{( 4 x y ^{3} )}

x \to - 1 lim (4 x^{2} - 3 x + 2)