# differentiate below functions

1) y= (t^3+t)/(t^4-3)

2)f(x)=cos(8x+2)

3)f(x)= cos(sin(x^2))
asked May 4, 2013 in CALCULUS

1)

y = (t3 + t) / (t4 - 3)

Diferenciate each side with respective t

dy / dt = d / dt((t3 + t) / (t4 - 3))

Recall : Diferenciate formula d(u / v) = (vu' - uv') / v^2

substitute u = t3 + t , v = t4 - 3 , u' = 3t2 + 1 and v' = 4t3 in the above diferecial formula

dy / dt = [(t4 - 3)(3t2 + 1) - (t3 + t)(4t2)]  /  (t4 - 3)2

dy / dt = [3t6 + t4 - 9t2  - 3 - 4t5 + 4t3] / (t4 - 3)^2

dy / dt = [3t6 - 4t5 + t4 + 4t3 - 9t2  - 3] / (t4 - 3)^2.

dy/dt = [(t4 - 3)(3t2 + 1) - (t3 + t)(4t3)]  /  (t4 - 3)2

dy/dt = [3t6 - 9t2 + t4 - 3 - (4t6 + 4t4)]  /(t4 - 3)2

dy/dt = [3t6 - 9t2 + t4 - 3 - 4t6 - 4t4)]  /(t4 - 3)2

dy/dt = [- t6 - 9t2 - 3t4 - 3]  /(t4 - 3)2

dy/dt = - ( t6 + 3t4 + 9t2 + 3]  /(t4 - 3)2

2)

f '(x) = cos(8x + 2)

Diferenciate with respective x

Recall :Diferenciate formula d / dx(cos(ax + b)) = -asin(ax + b)

f '(x) = -8sin(8x + 2).

3)

f(x) = cos(sin(x2)

sin(x^2) = t then dt = cos(x2) 2xdx

f(x) = cos(t)

Diferenciate each side with respective t

f ' (x)dx = -sin(t) dt

Substitute t = sin(x2) and dt = cos(x2) 2xdx in the above f ' (x)

f ' (x) dx = -sin(sin(x2)) cos(x2) 2x dx

f '(x) = -sin(sin(x2)) cos(x2) 2x.