Question

evaluate \\( \frac{d}{dt} e^{-4t^2 + 3t + 7} = \\)

Explanation:

Step1: Identify the function type

We have a function \( y = e^{-4t^2 + 3t + 7} \), and we need to find its derivative with respect to \( t \). This is a composite function, so we'll use the chain rule. The chain rule states that if we have a function \( y = e^{u(t)} \), then \( \frac{dy}{dt}=e^{u(t)}\cdot u'(t) \), where \( u(t) \) is the inner function. Here, \( u(t)=-4t^2 + 3t + 7 \).

Step2: Differentiate the inner function \( u(t) \)

We differentiate \( u(t)=-4t^2 + 3t + 7 \) with respect to \( t \). Using the power rule \( \frac{d}{dt}(t^n)=nt^{n - 1} \) and the sum/difference rule for differentiation:

• The derivative of \( -4t^2 \) with respect to \( t \) is \( -4\times2t=-8t \) (by power rule: \( n = 2 \), so \( 2t^{2-1}=2t \), multiplied by -4 gives -8t).
• The derivative of \( 3t \) with respect to \( t \) is \( 3 \) (by power rule: \( n = 1 \), so \( 1t^{1 - 1}=1 \), multiplied by 3 gives 3).
• The derivative of the constant \( 7 \) with respect to \( t \) is \( 0 \).

So, \( u'(t)=\frac{d}{dt}(-4t^2 + 3t + 7)=-8t + 3+0=-8t + 3 \).

Step3: Apply the chain rule

Now, using the chain rule \( \frac{dy}{dt}=e^{u(t)}\cdot u'(t) \), substitute \( u(t)=-4t^2 + 3t + 7 \) and \( u'(t)=-8t + 3 \):

\( \frac{d}{dt}e^{-4t^2 + 3t + 7}=e^{-4t^2 + 3t + 7}\cdot(-8t + 3) \)

We can also write this as \( (-8t + 3)e^{-4t^2 + 3t + 7} \).

Answer:

\( (-8t + 3)e^{-4t^2 + 3t + 7} \) (or \( e^{-4t^2 + 3t + 7}(-8t + 3) \))