Question

Question:-

Find the slope of the tangent to the curved
$\\ \sf \: y = \pi e^{\sin(x+y)} -\dfrac{\pi}{4} at \: the \: point \: \bigg(\frac{π}{4},\frac{ 3π}{4} \bigg).$

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Answer 1

Step-by-step explanation:

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Answer 2

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

Given curve is

$\rm \: y = \pi e^{\sin(x+y)} -\dfrac{\pi}{4}$

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

$\rm \: \dfrac{d}{dx}y =\dfrac{d}{dx}\bigg[ \pi e^{\sin(x+y)} -\dfrac{\pi}{4}\bigg]$

We know,

$\boxed{\tt{ \dfrac{d}{dx} {e}^{x} = {e}^{x} \: }}$

and

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

So, using these results, we get

$\rm \: \dfrac{dy}{dx} = \pi e^{\sin(x+y)}\dfrac{d}{dx}sin(x + y)$

We know,

So, using this, we get

$\rm \: \dfrac{dy}{dx} = \pi e^{\sin(x+y)} \: cos(x + y) \: \dfrac{d}{dx}(x + y)$

$\rm \: \dfrac{dy}{dx} = \pi e^{\sin(x+y)} \: cos(x + y) \: \bigg(1 + \dfrac{dy}{dx}\bigg)$

can be rewritten as

$\rm \: \dfrac{dy}{dx}= \pi e^{\sin(x+y)} \: cos(x + y) \: + \pi e^{\sin(x+y)} \: cos(x + y) \dfrac{dy}{dx}$

$\rm \: \dfrac{dy}{dx} - \pi e^{\sin(x+y)} \: cos(x + y) \dfrac{dy}{dx} = \pi e^{\sin(x+y)} \: cos(x + y)$

$\rm \: \dfrac{dy}{dx}\bigg(1 - \pi e^{\sin(x+y)} \: cos(x + y)\bigg) = \pi e^{\sin(x+y)} \: cos(x + y)$

$\rm \: \dfrac{dy}{dx} = \dfrac{ \pi e^{\sin(x+y)} \: cos(x + y)}{1 - \pi e^{\sin(x+y)} \: cos(x + y)}$

So,

$\rm \: \dfrac{dy}{dx}_{\bigg(\dfrac{π}{4},\dfrac{ 3π}{4} \bigg)} = \dfrac{ \pi e^{\sin( \frac{\pi}{4} + \frac{3\pi}{4} )} \: cos( \frac{\pi}{4} + \frac{3\pi}{4} )}{1 - \pi e^{\sin( \frac{\pi}{4} + \frac{3\pi}{4} )} \: cos( \frac{\pi}{4} + \frac{3\pi}{4} )}$

$\rm \: \dfrac{dy}{dx}_{\bigg(\dfrac{π}{4},\dfrac{ 3π}{4} \bigg)} = \dfrac{ \pi e^{\sin\pi} \: cos( \pi)}{1 - \pi e^{\sin(\pi)} \: cos(\pi)}$

We know,

So, using this, we get

$\rm \: \dfrac{dy}{dx}_{\bigg(\dfrac{π}{4},\dfrac{ 3π}{4} \bigg)} = \dfrac{ \pi e^{0} \: ( - 1)}{1 - \pi e^{0} \: ( - 1)}$

$\rm \: \dfrac{dy}{dx}_{\bigg(\dfrac{π}{4},\dfrac{ 3π}{4} \bigg)} = \dfrac{ - \pi}{1 + \pi }$

Hence,

$\rm \: Slope\:of\:tangent \: = \: \dfrac{dy}{dx}_{\bigg(\dfrac{π}{4},\dfrac{ 3π}{4} \bigg)} = \dfrac{ - \pi}{1 + \pi }$

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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.

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.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$