PLEASE HELP!!!!!!

Suppose u and v are functions of x that are differentiable at x=0 and that u(0)= 7,u'(0)=-5,v(0)= -1, v'(0)= -4.

andriy [413]

This question is incomplete, the complete question is;

Suppose u and v are functions of x that are differentiable at x=0

and that { u(0) = 7, u'(0) = -5 } { v(0)= -1, v'(0) = -4 }

Find the values of the following derivatives at x = 0.

a) $\frac{d}{dx}$ ( uv )

b) $\frac{d}{dx}$ ( $\frac{u}{v}$ )

c) $\frac{d}{dx}$ ( $\frac{v}{u}$ )

Answer:

a) $\frac{d}{dx}$ ( uv ) = -23

b) $\frac{d}{dx}$ ( $\frac{u}{v}$ ) = 33

c) $\frac{d}{dx}$ ( $\frac{v}{u}$ ) = -32/49 or - 0.6531

Step-by-step explanation:

Given that;

{ u(0) = 7, u'(0) = -5 } { v(0)= -1, v'(0) = -4 }

a)

$\frac{d}{dx}$ ( uv )

we differentiate

$\frac{d}{dx}$ ( uv ) = uv' + vu'

at x = (0), we substitute our values

$\frac{d}{dx}$ ( uv ) = ( 7 × -4 ) + ( -1 × -5)

$\frac{d}{dx}$ ( uv ) = -28 + 5

$\frac{d}{dx}$ ( uv ) = -23

b)

$\frac{d}{dx}$ ( $\frac{u}{v}$ )

we differentiate

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = ( vu' - uv' ) / v²

at x=0, we substitute our values

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = ( (-1 × -5) - (7 × -4 ) ) / (-1)²

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = (( 5 - ( -28 )) / 1

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = 33 / 1

$\frac{d}{dx}$ ( $\frac{u}{v}$ ) = 33

c) $\frac{d}{dx}$ ( $\frac{v}{u}$ )

we differentiate

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( uv' - vu' ) / u²

at x=0, we substitute our values

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( (7 × -4) - (-1 × -4) ) / (7)²

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( -28 - ( 4 ) ) / 49

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = ( -28 - 4 ) /49

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = -32 / 49

$\frac{d}{dx}$ ( $\frac{v}{u}$ ) = -32/49 or - 0.6531

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