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How do you find dy/dx given #y=ln(2+x^2)#?

Answer 1

Samantha Adkison

#(2x)/(2+x^2)#

Using #d/dx(ln(f(x)))=1/f(x)#

combined with the #color(blue)" chain rule"#

#d/dx[f(g(x))]=f'(g(x)).g'(x)color(green)" A"# #"------------------------------------------------"#

here #f(g(x))=ln(2+x^2)rArrf'(g(x))=1/(2+x^2)#

and #g(x)=2+x^2rArrg'(x)=2x# #"---------------------------------------------------------------"# Substitute these values into#color(green)" A"#

#rArrdy/dx=1/(2+x^2) xx2x=(2x)/(2+x^2)#

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Answer 2

Emma Aldridge

To find ( \frac{dy}{dx} ) given ( y = \ln(2 + x^2) ), you apply the chain rule. The derivative of ( \ln(u) ) with respect to ( x ) is ( \frac{1}{u} \frac{du}{dx} ). In this case, ( u = 2 + x^2 ). Thus,

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

Using the power rule, ( \frac{d(2 + x^2)}{dx} = 2x ). Substituting back,

[ \frac{dy}{dx} = \frac{1}{2 + x^2} \cdot 2x = \frac{2x}{2 + x^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.