How do you find the derivative of #y=arcsin(5x+5)#?

Answer 1

#(dy)/(dx)=5/sqrt(1-(5x+5)^2)#

If #u=arcsinv# then #v=sinu#, hence #(dv)/(du)=cosu# and

#(du)/(dv)=1/cosu=1/sqrt(1-sin^2u)=1/sqrt(1-v^2)#

Hence, we can now find the the derivative of #y=arcsin(5x+5)# using the concept function of a function and this is given by

#(dy)/(dx)=1/sqrt(1-(5x+5)^2)xxd/(dx)(5x+5)# or

#(dy)/(dx)=5/sqrt(1-(5x+5)^2)#

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Answer 2

Emma Aldridge

To find the derivative of ( y = \arcsin(5x + 5) ), you can use the chain rule of differentiation. The derivative is:

[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - (5x + 5)^2}} \cdot (5) ]

So,

[ \frac{dy}{dx} = \frac{5}{\sqrt{1 - (5x + 5)^2}} ]

This is the derivative of ( y = \arcsin(5x + 5) ).

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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.