Question:

Consider the equation given by the implicit function:

a) Find $\frac{{dy}}{{dx}}$ using implicit differentiation.

b) Determine the slope of the tangent line to the curve at the point $(1,1)$.

c) Find the equation of the tangent line to the curve at the point $(1,1)$.

Answer:

a) To find $\frac{{dy}}{{dx}}$ using implicit differentiation, we differentiate both sides of the equation with respect to $x$ and apply the chain rule.

Given equation: $x^3 + 3xy^2 - 5x = 2y^3 - 6xy + 4$

Differentiating both sides with respect to $x$, we get:

Using the chain rule and product rule on the left side, and the chain rule on the right side, we obtain:

Rearranging the terms, we have:

Simplifying further, we obtain:

Finally, dividing both sides by $3x^2$, we get the expression for $\frac{{dy}}{{dx}}$:

b) To find the slope of the tangent line to the curve at the point $(1,1)$, we substitute $x = 1$ and $y = 1$ into the expression $\frac{{dy}}{{dx}}$ we obtained above.

Substituting $x = 1$ and $y = 1$ into $\frac{{dy}}{{dx}}$ gives:

Therefore, the slope of the tangent line to the curve at the point $(1,1)$ is $\frac{8}{3}$.

c) The equation of a line can be determined using the point-slope form $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line and $m$ is the slope of the line.

Using the point $(1,1)$ and the slope $\frac{8}{3}$, we substitute these values into the point-slope form equation:

Multiplying both sides by 3 to eliminate the fraction, we get:

Expanding and rearranging, we obtain the equation of the tangent line:

Therefore, the equation of the tangent line to the curve at the point $(1,1)$ is $8x - 3y - 5 = 0$.