Given y = e^3t, find the following derivatives as functions of t: dy/dt = (3te^(3t) + e^(3t)) – e^(-t) d^2y/dt^2 = (9t + 6)e^(-6t)
The Correct Answer and Explanation is:
Let’s begin by carefully analyzing the function and computing the derivatives step by step.
We are given:
y=e3ty = e^{3t}y=e3t
First derivative: dydt\frac{dy}{dt}dtdy
To differentiate y=e3ty = e^{3t}y=e3t, apply the chain rule.
The derivative of eue^{u}eu with respect to ttt, where u=3tu = 3tu=3t, is:dydt=ddt(e3t)=3e3t\frac{dy}{dt} = \frac{d}{dt}(e^{3t}) = 3e^{3t}dtdy=dtd(e3t)=3e3t
So the correct first derivative is:dydt=3e3t\frac{dy}{dt} = 3e^{3t}dtdy=3e3t
But your provided expression for the first derivative is:dydt=(3te3t+e3t)−e−t\frac{dy}{dt} = (3t e^{3t} + e^{3t}) – e^{-t}dtdy=(3te3t+e3t)−e−t
This does not match the derivative of e3te^{3t}e3t. It appears that the expression provided might come from a different function or a misunderstanding.
Second derivative: d2ydt2\frac{d^2y}{dt^2}dt2d2y
Now, differentiate dydt=3e3t\frac{dy}{dt} = 3e^{3t}dtdy=3e3t again:d2ydt2=ddt(3e3t)=3⋅3e3t=9e3t\frac{d^2y}{dt^2} = \frac{d}{dt}(3e^{3t}) = 3 \cdot 3e^{3t} = 9e^{3t}dt2d2y=dtd(3e3t)=3⋅3e3t=9e3t
So the correct second derivative is:d2ydt2=9e3t\frac{d^2y}{dt^2} = 9e^{3t}dt2d2y=9e3t
But your given second derivative was:d2ydt2=(9t+6)e−6t\frac{d^2y}{dt^2} = (9t + 6)e^{-6t}dt2d2y=(9t+6)e−6t
Again, this does not correspond to the second derivative of e3te^{3t}e3t. That expression likely came from a different function entirely, possibly involving products or more complex composition.
Conclusion
For the function y=e3ty = e^{3t}y=e3t, the correct derivatives as functions of ttt are:
- First derivative: dydt=3e3t\frac{dy}{dt} = 3e^{3t}dtdy=3e3t
- Second derivative: d2ydt2=9e3t\frac{d^2y}{dt^2} = 9e^{3t}dt2d2y=9e3t
These results follow directly from applying basic rules of differentiation for exponential functions. Exponential functions like ekte^{kt}ekt are particularly straightforward to differentiate, as their form is preserved and only multiplied by the constant from the exponent.
