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What is the derivative of #sin^2(lnx)#?

Answer 1

Christopher Allers

We'll need the chain rule here, which states that #(dy)/(dx)=(dy)/(du)(du)/(dv)(dv)/(dx)#

Renaming #u=sin(v)# and #v=lnx#, we can follow the rule. Let's derivate it step-by-step:

#(dy)/(du)=2u#

#(du)/(dv)=cosv#

#(dv)/(dx)=1/x#

Then:

#(dy)/(dx)=2ucosv(1/x)#

Substituting #u#:

#(dy)/(dx)=2sin(v)cos(v)(1/x)#

Substituting #v#:

#(dy)/(dx)=(2sin(lnx)cos(lnx))/x#

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Answer 2

Aurora Alvarado

To find the derivative of sin²(lnx), you can use the chain rule. Let u = ln(x). Then, sin²(lnx) becomes sin²(u).

Now, differentiate sin²(u) with respect to u, which is 2sin(u)cos(u) using the chain rule.

Next, differentiate u = ln(x) with respect to x, which is 1/x using the derivative of ln(x).

Finally, apply the chain rule to find the derivative of sin²(lnx) with respect to x:

dy/dx = (dy/du) * (du/dx) = 2sin(u)cos(u) * (1/x) = 2sin(ln(x))cos(ln(x))/x.

Therefore, the derivative of sin²(lnx) with respect to x is 2sin(ln(x))cos(ln(x))/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.