How do you take the derivative of #tan^ -1(3x^2)#?

Answer 1

Evelyn Agard

#(6x)/(1+9x^4)#

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

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

#dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"#

#"noting that "d/dx(tan^-1x)=1/(1+x^2)#

#d/dx(tan^-1(3x^2))#

#=1/(1+(3x^2)^2)xxd/dx(3x^2)#

#=(6x)/(1+9x^4)#

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Answer 2

Christopher Allers

To take the derivative of ( \tan^{-1}(3x^2) ), you use the chain rule. The derivative of ( \tan^{-1}(u) ) with respect to ( x ) is ( \frac{1}{1+u^2} \frac{du}{dx} ). Here, ( u = 3x^2 ). Therefore, the derivative of ( \tan^{-1}(3x^2) ) with respect to ( x ) is ( \frac{1}{1+(3x^2)^2} \times \frac{d}{dx}(3x^2) ). Differentiating ( 3x^2 ) with respect to ( x ) yields ( 6x ). So, the final answer is ( \frac{6x}{1+9x^4} ).

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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.