Derivative/Equation of Tangent (1 Viewer)

[GaDaMIt]

firstly im a prelim. student so sorry if my mathematics is not quite up to scratch

anyway the question ..

Find the equations of the tangent and normal at the point where x = 1 to:

a) y = (5x - 4)^4

ive got as far as deriving it (by chain rule) to...

dy/dx = 4(5x-4)^3 (5)
= 20(5x - 4)^3

ive also worked out the y coordinate is 1 when the x is one by subbing into the original equation..

i dont get how to work out the equation of the tangent because that gradient is too complicated to work with

y - 1 = 20(5x-4)^3 (x - 1)

 

[airie]

Sub x=1 into dy/dx=20(5x-4)^3 to get the gradient of the tangent at x=1, which turns out to be 20. Then, by using the point-gradient form (is that what it's called? lol I never remember those names ), you get y-1=20(x-1), which is just y=20x-19.

 

[risaka]

since y = (5x-4)^4
dy/dx = 4(5x-4)^3.5
= 20(5x-4)^3
since u know the x coordinate sub x=1 into y
therefore the x&y corodinate is (1,1)
at (1,1) dy/dx = 20
then use ----> y-y1 = m (x-x1) to find the equation of tangent
y-1 = 20 (x-1)
therefore y=20x-19

 

[GaDaMIt]

okay thanks for that.. im just having another issue.. same question, different equation..

Find the equations of the tangent and normal at the point where x = 1

y= sqrt (x - 2)

when i sub in the value of x = 1 .. i get y = sqrt (-1) which obviously doesn't exist.. so what am i doing wrong?

 

[onebytwo]

you have ti differentiate the equation first

 

[GaDaMIt]

yeh my bad solutions are .. "there are no solutions"

anyway.. ive come across another question that i cant do since

Find the values of a and b if the parabola y = a(x+b)^2 - 8 has tangent y=2x at the point P(4,8)

 

[rama_v]

yeh my bad solutions are .. "there are no solutions"

anyway.. ive come across another question that i cant do since

Find the values of a and b if the parabola y = a(x+b)^2 - 8 has tangent y=2x at the point P(4,8)

differentiate as you would normally, taking a and b as constants

so dy/dx = 2a(x+b) =m

and since the gradient of the tangent is 2 at (4,8) then subbing this into the above equation gives:

m= 2
so 2 = 2a(4+b)
or a(b+4) = 1

Secondly you know that the point (4, 8) lies on the parabola. Sub x = 4, y = 8 into the equation of the parabola and you will get another equation involving a and b. Simulataneously solve to get the values of a and b

 

[GaDaMIt]

yeh thanks another Q...

Show that if a polynomial F(x) can be written as a product f(x)=(x-a)^n q(x) of the polynomials (x-a)^n and q(x), where n>=2, then f'(x) can be written as a multiple of (x-a)^(n-1). What does this say about the shape of the curve near x=a?

EDIT: AND....

Establish the rule for differentiating a product y=uvw, where u, v and w are functions of x. Hence find the derivative of these functions, and the values of x where the tangent is horizontal

a) x^5 (x-1)^4 (x-2)^3

 

[ice ken]

wow nsw does that stuff in yr 11 lol. i jst started to learn that stuff here in adel in beginning of the yr.

 

[rama_v]

Establish the rule for differentiating a product y=uvw, where u, v and w are functions of x. Hence find the derivative of these functions, and the values of x where the tangent is horizontal

A nice way would be to use logarithmic differentiation. For example let y = uv, where u and v are functions of x. Then,

y = uv
ln y = ln (uv)
ln y = ln u + ln v
(1/y)*y' = (1/u)*u' + (1/v)*v'

y' = y[(1/u)*u' + (1/v)*v']
y' = uv[(1/u)*u' + (1/v)*v']
y' = u'v + uv' which is the product rule. You can use the same trick for the question where y = uvw.

 

[Anonymou5]

Just keep on moving the prime across. For example (uvw)' = u'vw + uv'w + uvw' and (uvwt)' = u'vwt + uv'wt + uvw't + uvwt'.