Question

If f (x) = 3+2sinx+5cosx/cosx, then f' (x) =

Answer 1

Answer:

f'(x) = 3secx•tanx + 2sec²x

Solution:

• GIVEN : f(x) = (3 + 2sinx + 5cosx)/cosx

• TO FIND : f'(x) = ?

We have ;

=> f(x) = (3 + 2sinx + 5cosx)/cosx

=> f(x) = 3/cosx + 2sinx/cosx + 5cosx/cosx

=> f(x) = 3secx + 2tanx + 5

Now,

Differentiating both sides wrt x , we get ;

=> df(x)/dx = d3secx/dx +d2tan/dx +d5/dx

=> f'(x) = 3•dsecx/dx + 2•dtan/dx + 0

=> f'(x) = 3secx•tanx + 2sec²x

Hence,

f'(x) = 3secx•tanx + 2sec²x