derivative of (2x+3^5+10x(2x+3)^4)

Lösung

\frac{d}{d x} ((2 x + 3)^{5} + 10 x (2 x + 3)^{4})

960 x^{4} + 4800 x^{3} + 8640 x^{2} + 6480 x + 1620

Schritte zur Lösung

\frac{d}{d x} ((2 x + 3)^{5} + 10 x (2 x + 3)^{4})

Vereinfache

(2 x + 3)^{5} + 10 x (2 x + 3)^{4} : 192 x^{5} + 1200 x^{4} + 2880 x^{3} + 3240 x^{2} + 1620 x + 243

= \frac{d}{d x} (192 x^{5} + 1200 x^{4} + 2880 x^{3} + 3240 x^{2} + 1620 x + 243)

Wende die Summen-/Differenzregel an:

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

= \frac{d}{d x} (192 x^{5}) + \frac{d}{d x} (1200 x^{4}) + \frac{d}{d x} (2880 x^{3}) + \frac{d}{d x} (3240 x^{2}) + \frac{d}{d x} (1620 x) + \frac{d}{d x} (243)

\frac{d}{d x} (192 x^{5}) = 960 x^{4}

\frac{d}{d x} (1200 x^{4}) = 4800 x^{3}

\frac{d}{d x} (2880 x^{3}) = 8640 x^{2}

\frac{d}{d x} (3240 x^{2}) = 6480 x

\frac{d}{d x} (1620 x) = 1620

\frac{d}{d x} (243) = 0

= 960 x^{4} + 4800 x^{3} + 8640 x^{2} + 6480 x + 1620 + 0

Vereinfache

= 960 x^{4} + 4800 x^{3} + 8640 x^{2} + 6480 x + 1620

21 x y^{^{'}} + y = 462 x

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