Next, we differentiate x with respect to t. We use the chain rule for differentiating u^a, where u = t + (1)/(t). The derivative is a u^(a-1) · (du)/(dt). So, we get a(t+(1)/(t))^(a-1) multiplied by the derivative of t + (1)/(t), which is 1 - (1)/(t²).

Next, we differentiate x with respect to t. We use the chain rule for differentiating u^a, where u = t + (1)/(t). The derivative is a u^(a-1) · (du)/(dt). So, we get a(t+(1)/(t))^(a-1) multiplied by the derivative of t + (1)/(t), which is 1 - (1)/(t²).

Since we need (dt)/(dx) for the chain rule, we take the reciprocal of (dx)/(dt).

Now we multiply (dy)/(dt) by (dt)/(dx) to find (dy)/(dx). We substitute the expressions we found in the previous steps.

We can cancel out the common term (1 - (1)/(t²)) from the numerator and the denominator, assuming t ≠ ± 1. This leaves us with the simplified expression for (dy)/(dx).

For a positive constant $a$ find $\frac{d y}{d x}$ , where $y=a^{i+\frac{1}{t}} \text {, and } x=\left(t+\frac{1}{t}\right)^a$

Answer: $\frac{d y}{d x} = \frac{a^{t+\frac{1}{t}} \ln a}{a\left(t+\frac{1}{t}\right)^{a-1}}$

Step-by-step solution

Step 1: Differentiate y with respect to t

We need to find $\frac{dy}{dx}$ . Since both $y$ and $x$ are functions of $t$ , we can use the chain rule: $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$ . First, let's differentiate $y$ with respect to $t$ . We use the formula for differentiating $b^{f(t)}$ , which is $b^{f(t)} \ln b \cdot f'(t)$ . Here, $b=a$ and $f(t) = t + \frac{1}{t}$ . The derivative of $t + \frac{1}{t}$ is $1 - \frac{1}{t^2}$ .

Step 2: Differentiate x with respect to t

Next, we differentiate $x$ with respect to $t$ . We use the chain rule for differentiating $u^a$ , where $u = t + \frac{1}{t}$ . The derivative is $a u^{a-1} \cdot \frac{du}{dt}$ . So, we get $a\left(t+\frac{1}{t}\right)^{a-1}$ multiplied by the derivative of $t + \frac{1}{t}$ , which is $1 - \frac{1}{t^2}$ .

Step 3: Find dt/dx

Since we need $\frac{dt}{dx}$ for the chain rule, we take the reciprocal of $\frac{dx}{dt}$ .

Step 4: Apply the Chain Rule

Now we multiply $\frac{dy}{dt}$ by $\frac{dt}{dx}$ to find $\frac{dy}{dx}$ . We substitute the expressions we found in the previous steps.

Step 5: Simplify the expression

We can cancel out the common term $\left(1 - \frac{1}{t^2}\right)$ from the numerator and the denominator, assuming $t \neq \pm 1$ . This leaves us with the simplified expression for $\frac{dy}{dx}$ .

More from this chapter

All Continuity and Differentiability solutions →

Solve your own maths question free →

For a positive constant aaa find dydx\frac{d y}{d x}dxdy​, where y=ai+1t, and x=(t+1t)ay=a^{i+\frac{1}{t}} \text {, and } x=\left(t+\frac{1}{t}\right)^ay=ai+t1​, and x=(t+t1​)a