- #1

- 2

- 0

product rule for the rest,

f' cos6

g' 1/x

f'g*g'f + (xy)<- part I get confused with..

- Thread starter PlusCrime
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- #1

- 2

- 0

product rule for the rest,

f' cos6

g' 1/x

f'g*g'f + (xy)<- part I get confused with..

- #2

Mark44

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Product rule and chain rule.xy.. product rule

No, you need the chain rule first, since there's a composite function, √(sin(6x) * ln(x)), with a product inside.product rule for the rest,

What is f' cos6 supposed to mean? If this is supposed to mean the derivative of cos(6x), it's wrong.f' cos6

g' 1/x

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.

f'g*g'f + (xy)<- part I get confused with..

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