If x=3t^-2 and y= e^-10t^2 compute dy/dx

[BigBeachBanana]

To expand on the previous post. Let's look at [imath]t^2[/imath] vs. [imath]2^{t}[/imath].
[imath]t^2[/imath] is a polynomial where your variable, t, is at the base and has a constant power of 2. In this case, you would use the power rule as you described "multiply by the power and then minus 1":
[math]=>\frac{d(\mathrm{t}^{n})}{dt} = nt^{n-1}[/math][math]=>\frac{d(\mathrm{t}^{2})}{dt} = 2t^{2-1}=2t[/math]Whereas [imath]2^{t}[/imath] is an exponential function where your variable, t, is in the exponent, and the base is a constant of 2. In this case, the power rule doesn't apply, you'd have to use the derivative of exponential functions:
[math]=>\frac{d(a^{t})}{dt}=a^{t}*ln(a)\\ =>\frac{d(2^{t})}{dt}=2^{t}*ln(2)[/math]Note: when you have base e, [imath]\frac{d(e^{t})}{dt}=e^{t}*ln(e)=e^{t}*1=e^{t}[/imath]
Now, go back to your OP question. First apply the exponential rule, but now you also have to apply the chain rule. Hence the [imath]\frac{d}{dt}(-10t^2)[/imath] term. But [imath]-10t^2[/imath] is an polynomial, so you would use the power rule:
[math]\frac{d}{dt}(\mathrm{e}^{-10t^2}) = \mathrm{e}^{-10t^2}* ln(e)*\frac{d}{dt}(-10t^2)=\mathrm{e}^{-10t^2}* 1*(-20t)[/math]Hope this clear up your confusion

 

[ugu]

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I think you're confused between the derivative of polynomials and exponentials. Let's look at simpler problems [imath]t^2[/imath] vs. [imath]e^{2t}[/imath].
[imath]t^2[/imath] is a polynomial where your variable t is at the base, whereas [imath]e^{2t}[/imath] is an exponential function where your t variable is in the exponent. Does this make sense?

probably