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

Answer 1

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

When separating a product of two functions, the rule to follow 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 the standard rules have been applied

#d/dx sin x = cos x, qquad d/dx cot x = -cot x csc x#
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Answer 2

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

According to the product rule,

#(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)#.

Having established the rules, let's calculate 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)#
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Answer 3

To differentiate ( f(x) = \sin(2x) \cdot \cot(5-x) ) using the product rule:

[ f'(x) = (\sin(2x))' \cdot \cot(5-x) + \sin(2x) \cdot (\cot(5-x))' ]

Apply the chain rule and derivative of cotangent:

[ f'(x) = (2\cos(2x)) \cdot \cot(5-x) + \sin(2x) \cdot \left( -\csc^2(5-x) \right) ]

So, the derivative of ( f(x) = \sin(2x) \cdot \cot(5-x) ) using the product rule is:

[ f'(x) = 2\cos(2x) \cdot \cot(5-x) - \sin(2x) \cdot \csc^2(5-x) ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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