Question

Find the slope of the tangent to the curve $y = \pi e^{\sin(x+y)} -\dfrac{\pi}{4}$ at the point (π/4, 3π/4). ​

Answer 1

• refer attachment for detailed answer

sin 3π/4 = 1/√2, cos 3π/4 = - 1/√2

e = 2.17

π = 3.14

slop = -1.44 convert yourself in p/q form

Answer 2

Question:-

Find the slope of the tangent to the curve

$\sf \large \: y = \pi e^{sin(x+y)} -\dfrac{\pi}{4} \: at \: the \: point ( \frac{\pi}{4}, \frac{3\pi}{4} ).$

Given:-

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

To Find:-

• The slope of the tangent to the curve.

Solution:-

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

Differentiating with respect to x we get,

$\sf \large \frac{dy}{dx} = \pi e {}^{sin(x + y)} cos(x + y)(1 + \frac{dy}{dx} )$

$\sf \large \Rightarrow \frac{dy}{dx}(1 - \pi e {}^{sin(x + y)} cos(x + y) = \pi e {}^{sin(x + y)} \times cos(x + y)$

$\sf \large \therefore \frac{dy}{dx} = \frac{\pi e {}^{sin(x + y)} cos(x + y)}{1 - \pi e {}^{sin(x + y)} cos(x + y)}$

$\sf \large \therefore \bigg[ \frac{dy}{dx} \bigg] _{( \frac{\pi}{4} , \frac{3\pi}{4}) } = \frac{\pi e {}^{sin\pi} cos\pi}{1 - \pi sin\pi \times cos\pi }$

$\sf \large = - \frac{\pi}{1} = - \pi$

Answer:-

${ \boxed{ \sf \large \red{\underline {So, Slope = – \: \pi.}}}}$

Hope you have satisfied. ⚘