Question

For the following exercises, find $dy/dx$ at the value of the parameter.$x=\sqrt{t},y=2t+4,t=9$

Short answer

$\frac{dy}{dx}=12$ at $t=9$.

Step-by-step solution

Differentiate x with respect to t

Given that $x=\sqrt{t}$, we apply the power rule for differentiation. The power rule states that if $x=t^n$, then $\frac{dx}{dt}=n\cdot t^{n-1}$. In this case, $x=t^{1/2}$, so:$$\frac{dx}{dt}=\frac{1}{2}{t}^{-1/2}=\frac{1}{2\sqrt{t}}$$.

Differentiate y with respect to t

For the equation $y=2t+4$, differentiate it normally:$$\frac{dy}{dt}=2$$This is because the derivative of a constant is 0, and the derivative of $2t$ with respect to $t$ is 2.

Find dy/dx using the chain rule

To find $\frac{dy}{dx}$, we use the chain rule which is $\frac{dy}{dx}=\frac{dy}{dt}÷\frac{dx}{dt}$. Substitute the derivatives into the chain rule: $$\frac{dy}{dx}=\frac{2}{1/(2\sqrt{t})}=2\times 2\sqrt{t}=4\sqrt{t}$$.

Evaluate dy/dx at t = 9

Substitute $t=9$ into $\frac{dy}{dx}=4\sqrt{t}$:$$\frac{dy}{dx}=4\sqrt{9}=4\times 3=12$$.

Key concepts

Understanding the Chain Rule in Calculus

The chain rule is a fundamental tool in calculus used for differentiating compositions of functions. When you have a function nested within another function, the chain rule simplifies finding their derivative by allowing you to compute the derivative of the outer function multiplied by the derivative of the inner function.
For example, say you have two functions where one is applied to the other, like in our exercise where the expressions involve both variable dependency and parameter relationships.
The chain rule states:

• If you have a composition of functions where one function is nested inside another, say functions $u$ and $v$, then the derivative of this composition $d(u(v(t)))/dt$ can be calculated as $(du/dv)\times (dv/dt)$.

In our exercise, to find $dy/dx$, we treated $x$ and $y$ as functions of $t$, a parameter. The chain rule allowed us to express $dy/dx$ as $dy/dt÷dx/dt$, which was crucial for solving the problem correctly.

Applying the Power Rule in Derivatives

The power rule is one of the simplest and most essential rules for differentiation, particularly useful when dealing with polynomial expressions.The power rule states that if you have a function of the form $f(t)=t^n$, the derivative of this function, $f^{\prime }(t)$, is $n\times t^{n-1}$.
This rule makes it much easier to differentiate terms with variables raised to a power.

• In the context of our exercise, we applied the power rule while differentiating $x=\sqrt{t}$, which can be rewritten as $x=t^{1/2}$.
• By applying the power rule, we derived $dx/dt=\frac{1}{2}{t}^{-1/2}=\frac{1}{2\sqrt{t}}$.

This simplified the process of finding the derivative by focusing directly on the exponent of the variable, emphasizing how easy the power rule makes deriving powers of $t$.

Calculating Derivatives Step-by-Step

Calculating derivatives involves systematically applying differentiation rules to find how a function changes concerning its variables.In our exercise, we engaged in a layer-by-layer approach to differentiate $x$ and $y$ each with respect to the parameter $t$.

• First, we differentiated $x=\sqrt{t}$ using the power rule, resulting in $dx/dt=\frac{1}{2\sqrt{t}}$.
• Next, we differentiated $y=2t+4$ directly, recognizing it as a straightforward linear function to get $dy/dt=2$.

After obtaining these individual derivatives, we used the chain rule to relate $dy/dx$. This process is crucial as it involves seeing how one variable's rate of change relates to another's in more complex systems.

Evaluating Functions and Their Derivatives

Function evaluation is the process of calculating the output of a function given an input value or parameter.In calculus, after computing a derivative, we often evaluate it at a specific point to find the instantaneous rate of change at that point.

• In our exercise, after determining $dy/dx$ in terms of $t$, we were tasked with evaluating it at $t=9$.
• Substituting $t=9$ into $dy/dx=4\sqrt{t}$, we calculated $4\sqrt{9}=12$, demonstrating that the rate of change of $y$ with respect to $x$ at this point is 12.

This step exemplifies how a derivative serves not just as a functional expression, but as a means to quantify changes in specific conditions of the variables involved.