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

Answer 1

Emilia Aguilar

# dy/dx = 5/sqrt(1-25x^2) #

# y = arcsin(5x) #

# siny = 5x # ..... [1]

We can now differentiate implicitly to get:

# cos(y)dy/dx = 5 # ..... [2]

Using the fundamental trig identity #sin^2A+cos^2A-=1# we can write:

# sin^2(y+cos^2(y)=1# # :. (5x)^2+cos^2(y)=1# (from [1]) # :. cos^2(y)=1-25x^2# # :. cos(y)=sqrt(1-25x^2)#

Substituting into [2] we get:

# sqrt(1-25x^2)dy/dx = 5 #

# :. dy/dx = 5/sqrt(1-25x^2) #

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Answer 2

Emma Aldridge

To find the derivative of (y = \arcsin(5x)), use the chain rule. The chain rule states that if you have a composite function (f(g(x))), then the derivative is (f'(g(x)) \cdot g'(x)).

The derivative of (\arcsin(x)) with respect to (x) is (\frac{1}{\sqrt{1-x^2}}).

For (y = \arcsin(5x)), (f(x) = \arcsin(x)) and (g(x) = 5x), so (f'(x) = \frac{1}{\sqrt{1-x^2}}) and (g'(x) = 5).

Applying the chain rule, we get: [ \frac{dy}{dx} = \frac{1}{\sqrt{1-(5x)^2}} \cdot 5 ] [ \frac{dy}{dx} = \frac{5}{\sqrt{1-25x^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.