a.How do you find the equation of A tangent line to the curve, y=tanx at given points (π/4, 1)?

Question

b.Using product rule twice to prove that f,g, and h are differentiable, then:
(fgh)'=f'gh+fg'h+fgh'

c. Find equation of a tangent line to the curve y=x+cosx at points (O,1)

d.If f(x)= 3x²+5x, find f'(2) and use it to find an equation of the tangent line to the parabola y=3x²-5x at the point (2,2)

Answer 1

a) y = tan(x), equation of tangent line at (pi/4, 1)

To find the equation of the tangent line at (pi/4, 1), we must first obtain the slope of the tangent line. This is found by solving for the first derivative.

dy/dx = sec^2(x)
The slope m at pi/4 is found by plugging in pi/4 for the derivative.

m = sec^2(pi/4) = 1/cos^2(pi/4) = 1/[sqrt(2)/2]^2 = 1/(2/4) =
4/2 = 2

To find the equation of the tangent line, use the slope formula
with (x1, y1) = (pi/4, 1) and (x2, y2) = (x, y).

(y - 1) / (x - [pi/4]) = 2
y - 1 = 2(x - [pi/4])
y - 1 = 2x - [pi/2]
y = 2x - [pi/2] + 1
y = 2x - [pi/2] + [2/2]
y = 2x + [2 - pi]/2

d) f(x) = 3x^2 + 5x
f'(x) = 6x + 5

m = 6(2) + 5 = 17

(y2 - y1)/(x2 - x1) = m
(y - 2)/(x - 2) = 17
y - 2 = 17(x - 2)
y - 2 = 17x - 34
y = 17x - 32

Answer 2

a) y=tanx => y' = sec^2 x
at (π/4, 1) , dy/dx = sec^2 (π/4) = 2
y = mx+c
y = 2x + c
1 = π/2 +c => c = - π/2 + 1

Equation of tangent ---
y = 2x - π/2 + 1

b) (fgh)'=f'gh+fg'h+fgh'
let g(h(x)) = u(x) or gh = u
fgh = fu
(fu)'=f'u+fu'
replacing u
(fgh)'= f'gh + f (gh)'
Using product rule for gh
(fgh)'= f'gh + f [gh' + g'h]
=> (fgh)'=f'gh+fg'h+fgh'
Proved

c) y=x+cosx
y' = 1 - sinx
at (0,1) y' = 1
y = mx +c be the eqn of tangent
y = x + c
1 = 0 + c

Answer y = x+1

d) f(x)= 3x²+5x, f'(2) = 17
y=3x²-5x y'(2) = 7
y = 7x + c
2 = 14 + c
=> c = -12
y = 7x -12

Answer 3

I will just help you with part A: All you have to do is differentiate the function to get the slope at the point you specified; next just use the point slope formula to get the rest.