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How do you find the derivative of #y = arcsin(3x + 2)#?

Answer 1

Christopher Allers

The answer is #=3/sqrt(1-(3x+2)^2)#

We need #cos^2x+sin2x=1#

and #(sinx)'=cosx#

#y=arcsin(3x+2)#

#:. siny=3x+2#

Differentiating with respect to #x#

#(siny)'=(3x+2)'#

#cosydy/dx=3#

#dy/dx=3/cosy#

#cos^2y=1-sin^2y=1-(3x+2)^2#

#cosy=sqrt(1-(3x+2)^2)#

So,

#dy/dx=3/sqrt(1-(3x+2)^2)#

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Answer 2

Charles Arocha

To find the derivative of ( y = \arcsin(3x + 2) ), we use the chain rule.

[ \frac{d}{dx}(\arcsin(u)) = \frac{1}{\sqrt{1-u^2}}\frac{du}{dx} ]

Let ( u = 3x + 2 ).

[ \frac{du}{dx} = 3 ]

Now, substitute ( u = 3x + 2 ) and ( \frac{du}{dx} = 3 ) into the chain rule formula.

[ \frac{d}{dx}(\arcsin(3x + 2)) = \frac{1}{\sqrt{1 - (3x + 2)^2}} \cdot 3 ]

Simplify if necessary, and that's the derivative of ( y = \arcsin(3x + 2) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.