Differentiation of cos?

Question

f(x) = [cos (x^2 + 1)] / x^3

f'(x) = -sin(x^2 + 1) * 2x

I'm not too sure what to do with the denominator, any help would be appreciated.

Answer 1

Differentiate the first function by applying this rule

diff(u / v) = (diff(u) * v - diff(v) * u ) / (v ^ 2)

where u and v are differentiable continuous functions.

diff() means d() / dx,

1.) diff(f(x)) = diff[ cos ( x^2 + 1) ) * x^3 - diff ( x ^ 3) * cos ( x ^2 + 1) ] / (x ^ 6)

Now simplifying, we will get d(f(x)) / dx = (-3*cos(1 + x^2)) / x^4 - (2*sin(1 + x^2)) / x^2

2.) Differentiate it using the chain rule,

diff(f(x)) = diff( - sin(x^2 + 1) ) * 2x + diff(x^3) * (-sin(x^2 + 1))

diff(f(x)) = -4*x^2*cos(1 + x^2) - 2*sin(1 + x^2) [after simplifying]

Answer 2

Try this link, it may help.

    http://www.mathsrevision.net/alevel/page...

Answer 3

you have to use the quotient rule (or product rule if you prefer).

so breaking it down:

f(x) = g(x) / h(x)

   where g(x)=cos(x^2 + 1) and h(x)=x^3

f'(x) = [g'(x)*h(x) - g(x)*h'(x)] / h(x)^2

and then, you substitute in g(x), h(x), h'(x) = 3x^2, and finally g'(x)=-sin(x^2+1)*2x

Answer 4

f(x) = [cos (x² + 1)] / x³

f(x) = [cos (x² + 1)]  * (1/x³)

f'(x) = [cos (x² + 1)]'  * (1/x³) + [cos (x² + 1)]  * (1/x³)'

f'(x) = 2x (-sin (x² + 1))* (1/x³) + cos (x² + 1)  * (-3/x^4)'