A few derivative questions

  • #1
stonecoldgen
107
0
10. y=sin(lnx) find y'

im not sure how to do this, the previews question was actually with y=ln(sinx), which i know how to do, but not this one.

I know that the derivative of lnx is 1/x and the derivative of sinx=cosx



16.y=10[(x)2-sinx] find y'

PD: that is 10 to the power of (x2-sinx)



i know i should put a ln at both sides of the equation to use logarithmic differentiation, i know the logarithm properties. However, i don't know how should i apply all of that exactly.



EDIT: i also need help finding y' in y=x3-ln(x/e)


Thanks.
 

Answers and Replies

  • #2
Ignea_unda
133
0
The questions are very similar. The rule is known as the chain rule. The best way to describe is that you chain your way from the outside to the inside. If you know how to find y' for:
[itex]y = ln(sin(x))[/itex]

Then you know how to find y' for:
[itex] y = sin(ln(x))[/itex]

For the chain rule, if you have:
[itex] u = f(g(x))[/itex]
[itex] \frac{du}{dx} = \frac{df}{dg} \frac{dg}{dx} [/itex]

Does this help at all?
 
  • #3
stonecoldgen
107
0
The questions are very similar. The rule is known as the chain rule. The best way to describe is that you chain your way from the outside to the inside. If you know how to find y' for:
[itex]y = ln(sin(x))[/itex]

Then you know how to find y' for:
[itex] y = sin(ln(x))[/itex]

For the chain rule, if you have:
[itex] u = f(g(x))[/itex]
[itex] \frac{du}{dx} = \frac{df}{dg} \frac{dg}{dx} [/itex]

Does this help at all?


well yeah, all of the qustions kinda looked to chain rules like me, maybe not so ''strongly'' but they did.

Thanks.
 
  • #4
Ignea_unda
133
0
No worries. Sometimes it just takes a fresh glance to see it.

Glad I could help.
 

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