Find an equation of the tangent line to the curve at the given point. y = (4x is

Question

Find an equation of the tangent line to the curve at the given point.

y =

(4x is over x+1)

,    (3, 3)

y =

Differentiate.

f(x) = 6

sin x

Differentiate.

g(t) = 2 sec t + 7 tan t

Differentiate.

y =

(5-secx is over tanx)

y' =

Differentiate.

y = 4x2 sin x tan x

y' =

Find the derivative of the function.

F(x) = (x4 + 5x2 ? 7)9

F'(x)

Find the derivative of the function.

f(t) =

f(t) =

Find the derivative of the function.

f(x) = (2x ? 8)4(x2 + x + 1)5

Find the derivative of the function.

y =

Find the derivative of the function.

f(t) =

(t is over t^2 +6)

Explanation / Answer

1) y = 4x/(x+1)

dy/dx = d(4x/x+1)/dx = x+1*d(4x) - 4x(d(x+1))/(x+1^2) = 4x+4-4x/(x+1^2) = 4/(x+1^2) at (3,3) = 4/(3+1^2) = 1/4

therefore the slope of tangent is 1/4

equation of tangent with slope 1/4 and passing through point (3,3) is :-

y-3 = 1/4*(x-3) = 4y = x+9

2) y = 6*sqrt(x)*sinx

dy/dx = d(6*sqrt(x)*sinx/dx = 6*1/2 *1/(sqrt(x)*(sinx) + 6*sqrt(x)*cosx

3) g(t) =  2 sec t + 7 tan t

dg(t)/dt = 2d(sect)/dt + 7d(tant)/dt = 2*sec t*tan t + (7*(tan t)^2)

4) y = (5- sec x)/tan x = 5*cot x - cosec x

dy/dx = 5*d(cot x)/dx - d(cosec x)/dx = 5*(-cosec x)^2 - (-cot x*cosec x) = -5+cos x /(sin x^2)

5) y = 4x2 sin x tan x

dy/dx = sin x*tan x*d(4x2) / dx + 4x2* sin x d(tan x)/dx + 4x2*tan x*d(sin x)/dx

= sin x *tan x *8x + 4x2*sinx *(sec2x) + 4x2*tan x*cos x

6) F(x) = (x4 + 5x2 - 7)9

d(F(x))/dx = d( (x4 + 5x2 ? 7)9/dx = 9*(x4 + 5x2 - 7)8*(4x^3 +10x)

7) y = (2+tant)^1/6

dy/dt = d((2+tant)^1/6)/dt = 1/6*(2+tant)^(-5/6)*(sec^2 x)

8) f(x) = (2x - 8)4(x2 + x + 1)5

d(f(x)/dx = (x2 + x + 1)5*(4*(2x - 8)3*2) + (2x - 8)4*5*(x2 + x + 1)4*(2x+1)

9) f(x) = ((x^2+4)/(x^2-4))^5

f'(x) = 5*((x^2+4)/(x^2-4))^4*(x^2-4*(2x) -(x^2+4*(2x)))/((x^2-4)^2) = 5*((x^2+4)/(x^2-4))^4*((-16x)/((x^2-4)^2)

10) f(t) = sqrt(t/(t2 +6)

f'(t) = 1/2*(1/sqrt(t/(t2 +6)) * [((t2+6)*1 - t*(2t)) / (t2 +6)2]

f'(t) =  1/2*(1/sqrt(t/(t2 +6))*[(6 - t2 / (t2 +6)2]

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