derivative y=ln(3x+1)e^{x^2+4}

Lösung

ableitung von

y = ln (3 x + 1) e^{x^{2} + 4}

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

Schritte zur Lösung

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

Wende die Produktregel an:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = ln (3 x + 1), g = e^{x^{2} + 4}

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

\frac{d}{d x} (ln (3 x + 1)) = \frac{3}{3 x + 1}

\frac{d}{d x} (e^{x^{2} + 4}) = e^{x^{2} + 4} \cdot 2 x

= \frac{3}{3 x + 1} e^{x^{2} + 4} + e^{x^{2} + 4} \cdot 2 x ln (3 x + 1)

Multipliziere Brüche:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{3 e ^{x^{2} + 4}}{3 x + 1} + 2 e^{x^{2} + 4} x ln (3 x + 1)

\int \frac{4 x ^{2} - 3 x + 2}{4 x ^{2} - 4 x + 3} d x

\int_{1}^{2} \frac{e ^{\frac{1}{x}}}{x ^{2}} d x

f a c t or 2 cos (x) + cos^{2} (x)

y^{^{''}} + 3 y^{^{'}} + 2 y = 6

a re a f (x) = cos (x), 0, 2 π