find an equation of the tangent line to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$ at the…

Explanation:

Step1: Differentiate the hyperbola equation implicitly

Differentiate $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}} = 1$ with respect to $x$. Using the power - rule, we have $\frac{2x}{a^{2}}-\frac{2y\cdot y'}{b^{2}}=0$.

Step2: Solve for $y'$

Rearrange the equation $\frac{2x}{a^{2}}-\frac{2y\cdot y'}{b^{2}}=0$ to find the slope of the tangent line. First, move $\frac{2x}{a^{2}}$ to the other side: $-\frac{2y\cdot y'}{b^{2}}=-\frac{2x}{a^{2}}$. Then, multiply both sides by $-\frac{b^{2}}{2y}$ (assuming $y\neq0$) to get $y'=\frac{b^{2}x}{a^{2}y}$.

Step3: Find the slope at the point $(x_0,y_0)$

Substitute $x = x_0$ and $y = y_0$ into the derivative. The slope $m$ of the tangent line at the point $(x_0,y_0)$ is $m=\frac{b^{2}x_0}{a^{2}y_0}$.

Step4: Use the point - slope form of a line

The point - slope form of a line is $y - y_1=m(x - x_1)$. Here, $x_1=x_0$, $y_1=y_0$ and $m=\frac{b^{2}x_0}{a^{2}y_0}$. So $y - y_0=\frac{b^{2}x_0}{a^{2}y_0}(x - x_0)$. Rearrange to get $y=\frac{b^{2}x_0}{a^{2}y_0}(x - x_0)+y_0$.

Answer:

$y=\frac{b^{2}x_0}{a^{2}y_0}(x - x_0)+y_0$

Explanation for the second part:

Step1: Apply the chain rule

Let $u = \tan^{-1}(7x)$. Then $y = u^{2}$. The derivative of $y$ with respect to $u$ is $\frac{dy}{du}=2u$ by the power - rule. The derivative of $u=\tan^{-1}(7x)$ with respect to $x$ is $\frac{du}{dx}=\frac{7}{1+(7x)^{2}}$ using the formula $\frac{d}{dx}\tan^{-1}(v)=\frac{v'}{1 + v^{2}}$ where $v = 7x$ and $v'=7$.

Step2: Apply the chain rule formula $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $\frac{dy}{du}=2u$ and $\frac{du}{dx}=\frac{7}{1+(7x)^{2}}$ into the chain - rule formula. Since $u=\tan^{-1}(7x)$, we have $\frac{dy}{dx}=2\tan^{-1}(7x)\cdot\frac{7}{1 + 49x^{2}}$.

Answer:

$\frac{14\tan^{-1}(7x)}{1 + 49x^{2}}$