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How do you find the derivative of #g(t)=e^(-3/t^2)#?

Answer 1

Elijah Abram

#g'(t)=(6e^(-3/t^2))/(t^3)#

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

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

#g'(t)=f'(h(t))xxh'(t)larrcolor(blue)"chain rule"#

#g(t)=e^(-3/t^2)#

#rArrg'(t)=e^(-3/t^2)xxd/dt(-3/t^2)#

#d/dt(-3/t^2)=d/dt(-3t^-2)#

#=6t^-3=6/t^3#

#rArrg'(t)=e^(-3/t^2)xx6/t^3#

#color(white)(rArrg'(t))=(6e^(-3/t^2))/t^3#

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Answer 2

Emilia Aguilar

To find the derivative of ( g(t) = e^{-3/t^2} ), we can use the chain rule. The chain rule states that if ( f(x) ) and ( g(x) ) are differentiable functions, then the derivative of the composition ( f(g(x)) ) is given by ( f'(g(x)) \cdot g'(x) ).

In our case, ( f(x) = e^x ) and ( g(x) = -3/x^2 ). So, ( f'(x) = e^x ) and ( g'(x) = 6/x^3 ).

Now applying the chain rule, we have:

[ g'(t) = f'(g(t)) \cdot g'(t) = e^{-3/t^2} \cdot \left(\frac{d}{dt} \left(-\frac{3}{t^2}\right)\right) ]

[ g'(t) = e^{-3/t^2} \cdot \left(\frac{6}{t^3}\right) = \frac{6}{t^3} \cdot e^{-3/t^2} ]

So, the derivative of ( g(t) ) with respect to ( t ) is ( \frac{6}{t^3} \cdot e^{-3/t^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.