derivative of x^4+4x^2(y-2+5(y-1)^2)

Lösung

\frac{d}{d x} (x^{4} + 4 x^{2} (y - 2) + 5 (y - 1)^{2})

4 x^{3} + 4 (2 x (y - 2) + \frac{d y}{d x} x^{2}) + 10 (y - 1) \frac{d y}{d x}

Schritte zur Lösung

\frac{d}{d x} (x^{4} + 4 x^{2} (y - 2) + 5 (y - 1)^{2})

Wende die Summen-/Differenzregel an:

(f \pm g)^{'} = f^{'} \pm g^{'}

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

\frac{d}{d x} (x^{4}) = 4 x^{3}

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

\frac{d}{d x} (5 (y - 1)^{2}) = 10 (y - 1) \frac{d y}{d x}

= 4 x^{3} + 4 (2 x (y - 2) + \frac{d y}{d x} x^{2}) + 10 (y - 1) \frac{d y}{d x}

\frac{d}{d x} (x + in (x^{2} - 1))

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

d er i v a t i v e f (x) = \frac{2 x + 8 x ^{2}}{( 1 + 8 x ) ^{2}}

n = 0 \sum \infty \frac{1}{n ^{1}}

y^{^{''}} + 8 y^{^{'}} + 15 y = 4