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

Answer 1

Charles Arocha

#f'(x)=(5x)/(sqrt(1-x^2))+5arcsinx#

#"differentiate using the "color(blue)"product rule"#

#"given "f(x)=g(x)h(x)" then"#

#f'(x)=g(x)h'(x)+h(x)g'(x)larr" product rule"#

#g(x)=5xrArrg'(x)=5#

#h(x)=arcsinxrArrh'(x)=1/(sqrt(1-x^2))#

#rArrf'(x)=(5x)/(sqrt(1-x^2))+5arcsinx#

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Answer 2

Ezra Adams

To find the derivative of (f(x) = 5x \arcsin(x)), we will use the product rule. The derivative of (5x) is (5) and the derivative of (\arcsin(x)) is (\frac{1}{\sqrt{1-x^2}}). Applying the product rule, we get:

[f'(x) = 5 \cdot \frac{d}{dx}(\arcsin(x)) + \arcsin(x) \cdot \frac{d}{dx}(5x)]

[f'(x) = 5 \cdot \frac{1}{\sqrt{1-x^2}} + \arcsin(x) \cdot 5]

[f'(x) = 5 \cdot \frac{1}{\sqrt{1-x^2}} + 5\arcsin(x)]

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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.