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Assoli18 [71]
2 years ago
7

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.

Mathematics
1 answer:
andriy [413]2 years ago
6 0

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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