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differentiating inverse trig (1 Viewer)

Z

zinc

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how do you differentiate
y=tan-1(1/x) and y=1/(sin-1x)
thanks
 
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independantz

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y=arctan(1/x)
y'=1/(1+(1/x)^2)


y=1/arcsinx=(arcsinx)^-1
y'=(-1)(arcsinx)^-2 x(1/sqrt(1-x^2)), by the chain rule.
=-1/(sqrt(1-x^2))(arcsinx)^2
 
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vds700

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independantz said:
y=arctan(1/x)
y'=1/(1+(1/x)^2)


y=1/arcsinx=(arcsinx)^-1
y'=(-1)(arcsinx)^-2 x(1/sqrt(1-x^2)), by the chain rule.
=-1/(sqrt(1-x^2))(arcsinx)^2
Click to expand...
the first one is wrong...

y = arctan(1/x) , let u = 1/x = x^-1
dy/dx = dy/du . du/dx
=1/(1 + u^2). -x^-2
= -1/{[1 + (1/x^2)] . x^2}
= -1/(x^2 + 1)
 
Last edited:
lyounamu

lyounamu

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vds700 said:
the first one is wrong...

y = arctan(1/x) , let u = 1/x = x^-1
dy/dx = dy/du . du/dx
=1/(1 + u^2). -x^-2
= -1/{[1 + (1/x^2)] . x^2}
= -1/(x^2 + 1)
Click to expand...
Yeah, I think Andrew is right based on his working out.
 
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