How do you differentiate #f(x)=sin2x * cot(5-x)# using the product rule?

2 Answers
Mar 29, 2018

Answer:

# 2 cos(2x) cot(5-2x)+sin (2x)cot(5-x)csc(5-x)#

Explanation:

The rule for differentiating a product of two functions is

#d/dx (uv) = (du)/dx v+u (dv)/dx#

Thus

#f(x)=sin2x * cot(5-x) qquad implies qquad#

#d/dx f(x)=d/dx (sin2x )* cot(5-x) + sin2x * d/dx(cot(5-x))#
#qquad = (2 cos(2x) )* cot(5-2x)+sin (2x)*(-cot(5-x)csc(5-x)times (-1))#
# qquad = 2 cos(2x) cot(5-2x)+sin (2x)cot(5-x)csc(5-x)#

where we have used the standard rules

#d/dx sin x = cos x, qquad d/dx cot x = -cot x csc x#

Mar 29, 2018

Answer:

The derivative of #sin2xcot(5-x)# with respect to #x# is
#2cos2xcot(5-x)+sin2xcsc^2(5-x)#.

Explanation:

The product rule states

#(d[f(x)*g(x)])/(dx)=(df(x))/(dx)g(x)+f(x)(dg(x))/(dx)#

Here, #f(x)=sin2x#, and #g(x)=cot(5-x)#

Note that #f(x)# and #g(x)# are both COMPOSITE functions, so to solve this problem we will also need to use the chain rule which states that

#(df(g(x)))/(dx)=f^'(g(x))*g^'(x)#.

Now that we know the rules, let's compute the derivative

#(dsin2x*cot(5-x))/(dx)=(dsin2x)/(dx)cot(5-x)+sin2x(dcot(5-x))/(dx)#

#=2cos2xcot(5-x)+sin2x[-csc^2(5-x)*(-1)]#

#=2cos2xcot(5-x)+sin2xcsc^2(5-x)#