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How do you differentiate #5cos (2x)+ tan(3x+4)#?

Answer 1

Cameron Alexander

I would use the Chain Rule in both terms as: #y'=-5sin(2x)*2+1/(cos^2(3x+4))*3=# #=-10sin(2x)+3/(cos^2(3x+4))#

Where the Chain Rule allows you to derive #y=f(g(x))# as: #y'=f'(g(x))*g'(x)#

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Answer 2

Luke Alvarez

Let #f(u(x)) = 5cos(2x) + tan(3x+4)#.

For this, you would have to remember the following two derivatives:

#(d/(dx))[cosu] = -sinu(du(x))/(dx)# #(d/(dx))[tanu] = sec^2u(du(x))/(dx)#

since as you can see, #(df(u(x)))/(dx) = (df(u(x)))/cancel(du(x))*cancel(du(x))/(dx)#

where #u(x) = 2x# or #u(x) = 3x + 4#, in this case, for example. That just explains why the chain rule works how it works.

So, we get:

#(d/(dx))[5cos(2x) + tan(3x+4)]#

#= 5*-sin(2x)*2 + sec^2(3x+4)*3#

#= -10sin(2x) + 3sec^2(3x+4)#

Other ways you might see the chain rule:

In "prime" notation:

#[f(u(x))]' = f'(u(x))*u'(x)#

or in "function composition" notation:

#[f(x)@u(x)]' = [f'(x)@u(x)]*u'(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.