Thread: Trig

If f(x)=$\displaystyle cos^2x+cos^2(\frac{\pi}{3}+x)-cosxcos$($\displaystyle \frac{\pi}{3}+x)$ Find out the f'(x). In this sum How can I Easily derivate By substituting Method like $\displaystyle a=\frac{\pi}{3}$

So you would like to find out the following

d/dx(cos^2(x)+cos^2(pi/3+x)-cos(x) cos(pi/3+x))

Thus, Rewrite the expression as: cos^2(x)+cos^2(pi/3+x)-cos(x) cos(pi/3+x) = cos^2(x)-cos(x) sin(pi/6-x)+sin^2(pi/6-x)

= d/dx(cos^2(x)-cos(x) sin(pi/6-x)+sin^2(pi/6-x))

Differentiate the sum term by term and factor out constants:

= d/dx(cos^2(x))-d/dx(cos(x) sin(pi/6-x))+d/dx(sin^2(pi/6-x))

Using the chain rule, d/dx(cos^2(x)) = ( du^2)/( du) ( du)/( dx), where u = cos(x) and ( d)/( du)(u^2) = 2 u

= -(d/dx(cos(x) sin(pi/6-x)))+d/dx(sin^2(pi/6-x))+2 cos(x) d/dx(cos(x))

The derivative of cos(x) is -sin(x):

= -(d/dx(cos(x) sin(pi/6-x)))+d/dx(sin^2(pi/6-x))+-sin(x) 2 cos(x)

Simplify the expression:

= -(d/dx(cos(x) sin(pi/6-x)))+d/dx(sin^2(pi/6-x))-2 cos(x) sin(x)

Use the product rule:

d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x) and v = sin(pi/6-x):

= d/dx(sin^2(pi/6-x))-2 cos(x) sin(x)-cos(x) d/dx(sin(pi/6-x))+d/dx(cos(x)) sin(pi/6-x)

Simplify the expression:

= -(cos(x) (d/dx(sin(pi/6-x))))+d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)

Using the chain rule

d/dx(sin(pi/6-x)) = ( dsin(u))/( du) ( du)/( dx), where u = pi/6-x and ( d)/( du)(sin(u)) = cos(u):


= d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-cos(pi/6-x) d/dx(pi/6-x) cos(x)

Differentiate the sum term by term and factor out constants:

= d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-d/dx(pi/6)-d/dx(x) cos(pi/6-x) cos(x)

The derivative of pi/6 is zero:

= d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-cos(pi/6-x) cos(x) (-(d/dx(x))+0)

Simplify the expression:

= cos(pi/6-x) cos(x) (d/dx(x))+d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)

The derivative of x is 1:

= d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+1 cos(pi/6-x) cos(x)

Using the chain rule, d/dx(sin^2(pi/6-x)) = ( du^2)/( du) ( du)/( dx), where u = sin(pi/6-x) and ( d)/( du)(u^2) = 2 u:

= cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+2 d/dx(sin(pi/6-x)) sin(pi/6-x)

Using the chain rule, d/dx(sin(pi/6-x)) = ( dsin(u))/( du) ( du)/( dx), where u = pi/6-x and ( d)/( du)(sin(u)) = cos(u):

= cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+cos(pi/6-x) d/dx(pi/6-x) 2 sin(pi/6-x)

Differentiate the sum term by term and factor out constants:

= cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+d/dx(pi/6)-d/dx(x) 2 cos(pi/6-x) sin(pi/6-x)

The derivative of pi/6 again is zero:

= cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+2 cos(pi/6-x) sin(pi/6-x) (-(d/dx(x))+0)

Simplify the expression:

= cos(pi/6-x) cos(x)-2 cos(pi/6-x) (d/dx(x)) sin(pi/6-x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)

The derivative of x is 1:

= cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-1 2 cos(pi/6-x) sin(pi/6-x)

The derivative of cos(x) is -sin(x):

= cos(pi/6-x) cos(x)-2 cos(pi/6-x) sin(pi/6-x)-2 cos(x) sin(x)--sin(x) sin(pi/6-x)

Simplify the expression:

Thus, the answer:

= cos(pi/6-x) cos(x)-2 cos(pi/6-x) sin(pi/6-x)-2 cos(x) sin(x)+sin(pi/6-x) sin(x)