derivative of (4x-13ln((2x-7)(x-3)))

Lösung

\frac{d}{d x} ((4 x - 13) ln ((2 x - 7) (x - 3)))

4 ln ((2 x - 7) (x - 3)) + \frac{( 4 x - 13 ) ^{2}}{( 2 x - 7 ) ( x - 3 )}

Schritte zur Lösung

\frac{d}{d x} ((4 x - 13) ln ((2 x - 7) (x - 3)))

Wende die Produktregel an:

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

f = 4 x - 13, g = ln ((2 x - 7) (x - 3))

= \frac{d}{d x} (4 x - 13) ln ((2 x - 7) (x - 3)) + \frac{d}{d x} (ln ((2 x - 7) (x - 3))) (4 x - 13)

\frac{d}{d x} (4 x - 13) = 4

\frac{d}{d x} (ln ((2 x - 7) (x - 3))) = \frac{4 x - 13}{( 2 x - 7 ) ( x - 3 )}

= 4 ln ((2 x - 7) (x - 3)) + \frac{4 x - 13}{( 2 x - 7 ) ( x - 3 )} (4 x - 13)

\frac{4 x - 13}{( 2 x - 7 ) ( x - 3 )} (4 x - 13) = \frac{( 4 x - 13 ) ^{2}}{( 2 x - 7 ) ( x - 3 )}

= 4 ln ((2 x - 7) (x - 3)) + \frac{( 4 x - 13 ) ^{2}}{( 2 x - 7 ) ( x - 3 )}

\int_{- 2}^{3} (8 z^{3} + 3 z - 1) d z

\int_{1}^{\infty} \frac{( e ^{- x} )}{x} d x

y^{^{''}} - y^{^{'}} - 2 y = - 4 t + 8 t^{2}

\int y cos (z) d z

x y = (- 1 - x^{2}) y^{^{'}}