Question

find $\frac{dy}{dt}$ for each pair of functions. $y = x^2 - 4x$, $x = t^2 + 6$ $\frac{dy}{dt} = square$

Explanation:

Step1: Find dy/dx

Differentiate \( y = x^2 - 4x \) with respect to \( x \). Using the power rule, \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we get:
\( \frac{dy}{dx}=2x - 4 \)

Step2: Find dx/dt

Differentiate \( x = t^2+6 \) with respect to \( t \). Using the power rule, we have:
\( \frac{dx}{dt}=2t \)

Step3: Apply the chain rule

The chain rule states that \( \frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt} \). Substitute the values of \( \frac{dy}{dx} \) and \( \frac{dx}{dt} \) we found:
\( \frac{dy}{dt}=(2x - 4)\cdot(2t) \)
Now substitute \( x=t^2 + 6 \) into the expression:
\( \frac{dy}{dt}=(2(t^2 + 6)-4)\cdot(2t) \)
Simplify the expression inside the first parentheses: \( 2t^2+12 - 4=2t^2 + 8 \)
Then multiply by \( 2t \): \( (2t^2 + 8)\cdot(2t)=4t^3+16t \)

Answer:

\( 4t^3 + 16t \)