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How do you find the derivative of the function #y = arccos(e^(3x))#?

Answer 1

Cameron Alexander

#y'=(3e^(3x))/(sqrt(1-e^(6x))# Detail is as follows

#y=arccos(e^(3x))#.....(i) As If #y=arccosx# Then #y'=1/(sqrt(1-x^2))#......(ii)

Differentiating both sides of equation (i) and using equation (ii) #y'=1/sqrt(1-e^(6x)).e^(3x)(3)#

#y'=(3e^(3x))/sqrt(1-e^(6x))#

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Answer 2

William Abdalla

To find the derivative of the function ( y = \arccos(e^{3x}) ), you can use the chain rule. The derivative is:

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

Now, find the derivative of ( e^{3x} ), which is ( 3e^{3x} ). Substituting this back into the equation, you get:

[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - (e^{3x})^2}} \cdot 3e^{3x} ]

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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.