FIND THE EQUATION OF THE LINE TANGENT

Question

Answer 1

1)

a)

The equation f(x) = x(5x - 2)² and the x-coordinate of the point is 1.

Find the y-coordinate of the point.

y = (1)[5(1) - 2]² = 9.

Differentiate the equation with respect to x .

y' =x[2(5x - 2)(5)] + (5x - 2)²

Find slope of tangent line at the point P(1, 9).

m = (1)[2(5(1) - 2)(5)] + (5(1) - 2)² = 30 + 9 = 39.

Point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

y - (9) = 39(x - 1)

y - 9 = 39x -39

y = 39x - 30.

The tangent line equation is y = 39x - 30.

Comments

y = (1)[5(1) - 2]² = 9. Differentiate the equation with respect to x . y' =x[2(5x - 2)(5)] + (5x - 2)² how u get from that step to that step, and idk where the numbers are coming from?

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Given x- coordinate is 1

To find y -coordinate, substitute x = 1 is given given equation

f(x) = x(5x - 2)²

f(1) = 1(5(1) - 2)²

     =3^2

    =9

The given equation is f(x) = x(5x - 2)²

Differentiate the equation with respect to x .

Using product rule (uv)' = uv' +vu'

Here u = x and v = (5x - 2)²

u' = 1 and v' = 2(5x-2) (5x-2)' =  2(5x-2) (5) = 10(5x-2) 

f'(x) = x [10(5x-2)] + (5x - 2)²

      = 50x²-20x +25x²-20x +4

      = 75x²-40x +4

Slope at (1,9):

f'(x)  at  (1,9)   =  = 75(1)²-40(1) +4 =39

Answer 2

1)

b)

The equation f(x) = 4√(2x²+7) and the point is (3 ,20).

Differentiate the equation with respect to x .

y' =[4√(2x²+7) ]'

Let u = 2x²+7

u' = 4x

y' =4[√(u) ]'

y' =4[1/2√(u) ] u'                               [ derivative of √x = 1/2√x    ]

y' =4[1/2√(u) ] 4x

put u =  2x²+7

y' =16x/2√(2x²+7) 

y' =8x/√(2x²+7) 

Find slope of tangent line at the point P (3 ,20).

m = 8(3)/√(2(3)²+7) 

m = 24/√(18+7) 

m = 24/√(25) 

m = 24/5

Point-slope form of line equation : y - y₁ = m(x - x₁), where m = slope and (x₁, y₁) = point.

y - (20) = (24/5)(x - 3)

5(y - 20) = 24(x - 3)

5y - 100 = 24x - 72

5y = 24x+28

y = (24x+28)/5 .

The tangent line equation is y = (24x+28)/5 .

Comments

y' =4[1/2√(u) ] u' [ derivative of √x = 1/2√x ] how is derivatives of sqrt x is 1/2sqrtx?

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√x can be written as x^(1/2)

d/dx [x^(1/2) ]  = (1/2) x^(1/2  -1)         [ since derivative of  xⁿ = n x^n-1 ]

                        = (1/2) x^(-1/2 )     

                        = (1/2) [1/x^(1/2 )]       [ a^(-n) = 1/a^n   ]

                        =  (1/2)(1/√x )              [ x^(1/2 ) = √x  ]

                        = 1/2√x 

Hence derivative of √x = 1/2√x .