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# differentiation question watch

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1. how do i differentiate

y= (sqrt sinx)/(x²×tan³x)

is my working attached right?
Attached Images
2. calc.PDF (77.0 KB, 85 views)
3. Your working gets confusing at the fourth step. In any case, why insert ln at all? Differentiate it as a quotient...
y= (sqrt sinx)/(x²×tan³x)
y'=(x².tan³x.(cosx/2rt(sinx))-rt(sinx).(x².3tan²x.sec²x+2x. tan³x))/(x².tan³x)²
4. the attachment is working

ok surely when you took split the natural logs x sq and tan to the 3, you should have bracket around them as the negative sign applies to them both?
5. Or if you insist on ln...
lny= ln(sqrt sinx)/(x²×tan³x)
-->
lny=ln(rt(sinx))-ln(x².tan³x)
-->
y'/y=(1/rt(sinx)).(cosx/2rt(sinx)-(1/(x².tan³x)).(x².3tan²x.sec² x + 2x.tan³x)
-->
y'=(cosx/2sinx - (3x².sec²x + 2x.tanx)/(x².tanx)).y
-->
y'=(cosx/2sinx - (3x².sec²x + 2x.tanx)/(x².tanx)).(sqrt sinx)/(x²×tan³x)

More complicated, in my opinion.

Just don't forget to differentiate everything!
6. (Original post by J.F.N)
Or if you insist on ln...
lny= ln(sqrt sinx)/(x²×tan³x)
-->
lny=ln(rt(sinx))-ln(x².tan³x)
-->
y'/y=(1/rt(sinx)).(cosx/2rt(sinx)-(1/(x².tan³x)).(x².3tan²x.sec² x + 2x.tan³x)
-->
y'=(cosx/2sinx - (3x².sec²x + 2x.tanx)/(x².tanx)).y
-->
y'=(cosx/2sinx - (3x².sec²x + 2x.tanx)/(x².tanx)).(sqrt sinx)/(x²×tan³x)

More complicated, in my opinion.

Just don't forget to differentiate everything!
with this line
y'/y=(1/rt(sinx)).(cosx/2rt(sinx)-(1/(x².tan³x)).(x².3tan²x.sec² x + 2x.tan³x)
i can see what you have done but cant you use the log law logab = loga + logb instead of using the product rule?
7. (Original post by manps)
with this line
y'/y=(1/rt(sinx)).(cosx/2rt(sinx)-(1/(x².tan³x)).(x².3tan²x.sec² x + 2x.tan³x)
i can see what you have done but cant you use the log law logab = loga + logb instead of using the product rule?
Oh, you mean for ln(x².tan³x)? You can... But don't forget that its -(ln(x².tan³x))... so the term is -(ln(x²) + lntan³x))=-ln(x²)-ln(tan³x).

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