How Does the Product Rule Apply in Differentiating xy + sqrt[sin(6x)*ln(x)]?

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xy.. product rule

product rule for the rest,

f' cos6
g' 1/x

f'g*g'f + (xy)<- part I get confused with..
 
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PlusCrime said:
xy.. product rule
Product rule and chain rule.
PlusCrime said:
product rule for the rest,
No, you need the chain rule first, since there's a composite function, √(sin(6x) * ln(x)), with a product inside.
PlusCrime said:
f' cos6
g' 1/x
What is f' cos6 supposed to mean? If this is supposed to mean the derivative of cos(6x), it's wrong.

For the 2nd line, I get what you're doing, but you're not using the notation well.
d/dx(ln(x)) = 1/x
That gets across the idea that you're differentiating ln(x), and that the derivative is 1/x.

Another way to say it is, if g(x) = ln(x), then g'(x) = 1/x.
PlusCrime said:
f'g*g'f + (xy)<- part I get confused with..
 
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