Question

(1 point) Suppose y = 7x In r. Find the differential: dy = dr.

Answer 1

1) let

y = 7.rln (2)

Differentiate above equation

(NOTE:- product rule

)

U = uu' + vu'

day = (7.vlna

d day = 7cm (Ina) + (lnx) (72)

dcr y = - + (inx) 7

y = 7+ (in.) 7

dy = 17+7(In.c) d.c

(2)

Y = (2+6)

Differentiate above equation

(9+ z2) Rp

ay = 3(22 +6)24 (22 +6)

= 3(-² + 6)(2.r)

dy = (6.0(2 + 6)).

At x=4 and dx=0.05

dy = (6(4)(4² + 6))(0.05)

dy = (24(22))(0.05)

dy = 580.8

(3)

f(1) = -2.13 + 33.12 – 144.+2

differentiate above equation

(7 +2561 – + $47–) = = (2), f

f'() = -6.7% + 66.– 144

to get critical number equate

f'(x)=0

-62 + ᏮᏮᏆ - 1ᏎᏎ = 0

22 - 11.2 + 24 = 0

(1 – 3) 1 - 8) = 0

$x=3$ and $x=8$ are critical numbers.

(NOTE:- if f'(x)>0 then f(x) is increasing and if f'(x)<0 then f(x) is decreasing.)

Let f'(x)>0

-62 + ᏮᏮᏆ - 1ᏎᏎ >0

2-11-24<0

(1 – 3)(x -8) < 0

x belongs to

f(x) is increasing on

And decreasing on remaining interval

(-0,3) (8,0)

(4)

Clearly g(x) is increasing on

(3,0)

And
W() > 0
on
(-0, 0)

Hence h(x) is increasing on

(-0, 0)

(5)

Here g(x) is decreasing on

(-0,1)

And h'(x)<0 on

Hence h(x) is decreasing on ​​​​​​​