Question 167547: Hello all! I have this Trigonometry homework that is killing me and I need help from you guy's. Here's the problem let me know if you guys can solve it.
1. Calculate the derivatives:
I) I) (Sin4x) /( x² +1)
II) II) 10t²+1
III) III) √ ( 1 + u) / (1 – u)
IV) IV) (1) / ( x4 + 1)3
2. Calculate the points of the tangent of the curve : y = ex(x2 – 8) where and in which the tangent line is horizontal
3. Write one of the two tangents to the curve : y = 1 + x³ which are parallel to this equation 12x - y = 1.
4. Calculate
I) If f(x) = ex cos x calculate f ’(x) et f ’’
(ii) Calculate (d99/ dx99) Sin x
5.
Calculate the derivative :
I) y = xe-x²
II) Let say: r(x) = f(g(h(x))) whit h(1) = 2, g(2) = 3, h’(1)= 4, g’(2) = 5, f’(3) = 6. Calculate r’(1).
III) Show that the function : y = Ae-x + Be-x satisfy the equation with differential y”+ 2y’+y = 0
Answer by sowmya(32)   (Show Source):
You can put this solution on YOUR website! 1.Derivatives:
i)(Sin4x) /( x² +1)
Let y = (sin4x)/x^2 +1
dy/dx use u/v method
d(u/v) = (vdu-udv)/v^2
Here , u = sin4x
v= x^2 +1
du = 4 cos4x
dv = 2x
dy/dx = ((x^2 +1)4cos4x- sin4x(2x))/(x^2 +1)^2
= (4x^2 cos4x + 4 cos4x - 2x sin4x)/(x^2+1)^2
ii)10t^2 +1
y = 10 t^2 +1
dy/dt = 20 t
iii)y= sqrt((1+u)/(1-u))
y^2 = (1+u)/1-u)
2y dy = ((1-u) du + (1+u) du)/(1-u)^2
2y dy = 2du/(1-u)^2
y dy = du/(1-u)^2
dy/du = y/(1-u)^2
= sqrt((1-u)(1+u))/(1-u)^2
= (sqrt(1+u))/(1+u)^3/2
iv)question not clear
|
|
|
