Question

use implicit differentiation to find the slope of the tangent line to the curve 4x^2+2x+xy =2 at the point (2, -9)

Answer 1

$\large\underline{\sf{Solution-}}$

Let the point be P (2, - 9).

and

Given curve is

$\rm :\longmapsto\: {4x}^{2} + 2x + xy = 2$

On differentiating both sides, w. r. t. x, we get

$\rm :\longmapsto\: \dfrac{d}{dx}({4x}^{2} + 2x + xy) =\dfrac{d}{dx} 2$

We know,

$\red{ \boxed{ \sf{\dfrac{d}{dx}(u + v) \: = \: \dfrac{du}{dx} + \dfrac{dv}{dx}}}}$

and

$\red{ \boxed{ \sf{\dfrac{d}{dx}k\: = \: 0 }}}$

So, using these Identities, we get

$\rm :\longmapsto\: \dfrac{d}{dx}{4x}^{2} + \dfrac{d}{dx}2x +\dfrac{d}{dx} xy =0$

We know,

$\red{ \boxed{ \sf{\dfrac{d}{dx}k \: f(x)\: = \: k \: \dfrac{d}{dx} \: f(x) }}}$

and

$\red{ \boxed{ \sf{\dfrac{d}{dx} \: u.v \: = \: v\dfrac{d}{dx}u \: + \: u\dfrac{d}{dx}v }}}$

So, we get

$\rm :\longmapsto\:4 \dfrac{d}{dx}{x}^{2} +2 \dfrac{d}{dx}x +x\dfrac{d}{dx}y + y\dfrac{d}{dx}x =0$

We know,

$\red{ \boxed{ \sf{\dfrac{d}{dx} {x}^{n} \: = {nx}^{n - 1} }}}$

Using this,

$\rm :\longmapsto\:8x + 2 + x\dfrac{dy}{dx} + y = 0$

$\rm :\longmapsto\:x\dfrac{dy}{dx}= - 8x - 2 - y$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{ - 8x - y - 2}{x}$

Therefore,

$\red{ \boxed{ \sf{Slope \: of \: tangent \: = \: {\bigg(\dfrac{dy}{dx} \bigg) }_P}}}$

$\rm :\longmapsto\: \bigg(\dfrac{dy}{dx} \bigg)_{P} = \dfrac{ - 8 \times 2 - 2 + 9}{2}$

$\rm :\longmapsto\: \bigg(\dfrac{dy}{dx} \bigg)_{P} = \dfrac{ -16 + 7}{2}$

$\rm :\longmapsto\: \bigg(\dfrac{dy}{dx} \bigg)_{P} = \dfrac{ -9}{2}$

Hence,

$\red{ \boxed{ \sf{Slope \: of \: tangent \: = \: {\bigg(\dfrac{dy}{dx} \bigg) }_{P(2, - 9)}} = - \: \frac{9}{2} }}$

Additional Information :-

Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

$\red{ \boxed{ \sf{Slope \: of \: tangent \: = \: {\bigg(\dfrac{dy}{dx} \bigg) }_P}}}$

and

$\red{ \boxed{ \sf{Slope \: of \: normal\: = \frac{ - 1}{\bigg(\dfrac{dy}{dx} \bigg)_P} }}}$

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined

4. Two lines having slope M and m are parallel, iff M = m

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.