Question

Calculus help please. Attachment inside

Answer 1

i need help with 1b and 6

Answer 2

k

Answer 3

move the denominators to the top and make them negative exponents, then just use the normal power rule

Answer 4

x^-4 +5x^-5

Answer 5

\[\frac{ d }{ dx }[x ^{-4}] + 5*\frac{ d }{ dx }[x ^{-5}]\]

Answer 6

yep, and the derivative of a sum, is the sum of the derivatives

Answer 7

-4x^-5 -25x^-6

Answer 8

yep, not just fix those negative exponents on the final answer

Answer 9

okay thanks can we move onto 6 ?

Answer 10

sure

Answer 11

the derivative for 6 is 3x

Answer 12

The 'normal' will be perpendicular to the tangent at that point on the curve

Answer 13

y = x^3 + 1 at point (1,2) Find the Normal at the point.

Answer 14

\[\frac{ dy }{ dx } = 3x^2\]

Answer 15

okay do i plug in 1 ?

Answer 16

The derivative dy/dx tells you the slope of the tangent line at any point, so if you plug in x=1, the slope of the tangent is 3(1)^2 = 3

right

Answer 17

Do you remember what the slope of the line perpendicular to that tangent line will be?

Answer 18

?

Answer 19

Slope of the tangent at x=1 ....3
Slope of the normal at x=1, .....-1/3

The slope of the normal will be the negative reciprical of the slope of the tangent

Answer 20

oh ya

Answer 21

So just make a line equation with , slope m=-1/3 and through point (1,2)

Answer 22

y - y1 = m(x-x1)

Answer 23

y-2=-1/3(x-1)

Answer 24

yep, if you want to see

graph
y = x^3 + 1
and
the line of the tangent (slope = 3 ) point (1,2)
and
the line of the normal you just found

should be two perpendicular lines at that point

Answer 25

Here it is

Answer 26

Any other ones, or you good