How do you find the derivative of y=14tanxcosx+10cscx?

1 Answer

dy/dx=14cosx-10csc x cot x

Explanation:

In order to differentiate this function, we will need to use two different differentiation rules:

  1. The Chain Rule
  2. The Product Rule

Chain rule:

[f(x)]^n=n[f(x)]^(n-1)xxf'(x)

Product rule:

d/dxuv=u(dv)/dx+v(du)/dx

y=14tanxcosx + 10cscx

Let's split the function into two parts and differentiate each separately.

14tanxcosx

"Let" u = 14tanx "and" v=cosx

(du)/dx=14sec^2x

(dv)/dx=-sinx

d/dxuv=14cosx sec^2x-14tanxsinx

=14/cosx-(14sin^2x)/cosx=(14-14sin^2x)/cosx=(14(1-sin^2x))/cosx=(14cos^2x)/cosx=14cosx

d/dx10cscx=-10csc x cot x

dy/dx=14cosx-10csc x cot x

Also note that we can simplify the first expression:

d/dx 14 tanx cosx = d/dx 14 sinx/cosx cosx
" " = d/dx 14 sinx
" " = 14 cosx