Question

Use implicit differentiation to find the equation of the tangent line to the curve x * y ^ 3 + xy = 1 at the point (1, 1) . The equation of this tangent line can be written in the form y = mx + b where m is: and where b is:​

Answer 1

Step-by-step explanation:

We have,

$\rm \: x. {y}^{3} + xy = 1$

Differentiating both sides w.r.t x,

$\rm \implies \: {y}^{3} . \frac{d}{dx} (x) + x. \frac{d}{dx}( {y}^{3} ) + y. \frac{d}{dx}(x) + x \frac{d}{dx} (y) = 0$

$\rm \implies \: {y}^{3} . 1 + x.3 {y}^{2} \frac{dy}{dx} + y. 1 + x \frac{dy}{dx}= 0$

$\rm \implies \: {y}^{3} + 3x {y}^{2} \frac{dy}{dx} + y+ x \frac{dy}{dx}= 0$

$\rm \implies \: ({y}^{3} + y) + (3x {y}^{2} + x) \frac{dy}{dx}= 0$

$\rm \implies \: \frac{dy}{dx}= - \frac{{y}^{3} + y }{3x {y}^{2} + x}$

Now,

$\rm \implies \: \frac{dy}{dx} \bigg| _{(x = 1 , y = 1)} = - \frac{{(1)}^{3} + 1 }{3.(1) .{(1)}^{2} + 1}$

$\rm \implies \: \frac{dy}{dx} \bigg| _{(x = 1 , y = 1)} = - \frac{1 + 1 }{3.1 .1 + 1}$

$\rm \implies \: \frac{dy}{dx} \bigg| _{(x = 1 , y = 1)} = - \frac{2 }{3 + 1}$

$\rm \implies \: \frac{dy}{dx} \bigg| _{(x = 1 , y = 1)} = - \frac{2 }{4} = - \frac{1}{2}$

Equation of tangent:

$\rm(y - 1) = - \frac{ 1}{2} (x - 1)$

$\rm \implies \:y - 1= - \frac{ 1}{2}x + \frac{1}{2}$

$\rm \implies \:y = - \frac{ 1}{2}x + \frac{1}{2} + 1$

$\rm \implies \:y = - \frac{ 1}{2}x + \frac{3}{2}$

So, on comparing with $\rm\:y=mx+b$

we get, $\rm\:m=-\frac{1}{2}$ and$\rm\:b=\frac{3}{2}$