Question

Find the value of $\dfrac{dy}{dx}$ at $\theta =\dfrac{\pi }{4}$ , if $x=a{{e}^{\theta }}\left( \sin \theta -cos\theta \right)$ and $y=a{{e}^{\theta }}\left( \sin \theta +cos\theta \right)$ .

Answer 1

Hint: We know that $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}$ . So, find $\dfrac{dy}{d\theta }$ and $\dfrac{dx}{d\theta }$ separately and divide $\dfrac{dy}{d\theta }$ by $\dfrac{dx}{d\theta }$ to get the answer. For finding the derivative of x and y with respect to $\theta $ use the multiplication rule, i.e., uv rule of differentiation. Remember that derivative of sinx is cosx, derivative of cosx is –sinx and $\dfrac{d{{e}^{\theta }}}{d\theta }={{e}^{\theta }}$ .

Complete step by step answer:
Let us start the solution to the above question by finding $\dfrac{dy}{d\theta }$ . It is given that:
$y=a{{e}^{\theta }}\left( \sin \theta +cos\theta \right)$
If we differentiate both sides of the equation with respect to $\theta $ , we get
\[\dfrac{dy}{d\theta }=\dfrac{d\left( a{{e}^{\theta }}\left( \sin \theta +cos\theta \right) \right)}{d\theta }\]
Now, we will use the uv rule of differentiation. According to it $\dfrac{d\left( uv \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$ . So, we get
 \[\dfrac{dy}{d\theta }=a{{e}^{\theta }}\dfrac{d\left( \sin \theta +cos\theta \right)}{d\theta }+\left( \sin \theta +cos\theta \right)\dfrac{d\left( a{{e}^{\theta }} \right)}{d\theta }\]
Now, we know that that derivative of sinx is cosx, derivative of cosx is –sinx and $\dfrac{d{{e}^{\theta }}}{d\theta }={{e}^{\theta }}$ .
\[\dfrac{dy}{d\theta }=a{{e}^{\theta }}\left( \cos \theta -sin\theta \right)+\left( \sin \theta +cos\theta \right)a{{e}^{\theta }}\]
\[\Rightarrow \dfrac{dy}{d\theta }=2a{{e}^{\theta }}\cos \theta \]
Similarly, let us find the value of $\dfrac{dx}{d\theta }$ .
$x=a{{e}^{\theta }}\left( \sin \theta -cos\theta \right)$
If we differentiate both sides of the equation with respect to $\theta $ , we get
\[\dfrac{dx}{d\theta }=\dfrac{d\left( a{{e}^{\theta }}\left( \sin \theta -cos\theta \right) \right)}{d\theta }\]
Now, we will use the uv rule of differentiation.
 \[\dfrac{dx}{d\theta }=a{{e}^{\theta }}\dfrac{d\left( \sin \theta -cos\theta \right)}{d\theta }+\left( \sin \theta -cos\theta \right)\dfrac{d\left( a{{e}^{\theta }} \right)}{d\theta }\]
Now, we know that that derivative of sinx is cosx, derivative of cosx is –sinx and $\dfrac{d{{e}^{\theta }}}{d\theta }={{e}^{\theta }}$ .
\[\dfrac{dx}{d\theta }=a{{e}^{\theta }}\left( \cos \theta +sin\theta \right)+\left( \sin \theta -cos\theta \right)a{{e}^{\theta }}\]
\[\Rightarrow \dfrac{dx}{d\theta }=2a{{e}^{\theta }}\sin \theta \]
Now, let us move to find the value of $\dfrac{dy}{dx}$ . We know $\dfrac{dy}{dx}=\dfrac{\dfrac{dy}{d\theta }}{\dfrac{dx}{d\theta }}$ . So, if we substitute $\dfrac{dy}{d\theta }$ and $\dfrac{dx}{d\theta }$ from above results, we get
$\dfrac{dy}{dx}=\dfrac{2a{{e}^{\theta }}\cos \theta }{2a{{e}^{\theta }}\sin \theta }=\cot \theta $
Now, we will put $\theta =\dfrac{\pi }{4}$ . On doing so, we get
$\dfrac{dy}{dx}=\cot \dfrac{\pi }{4}=1$

So, the correct answer is 1.

Note: Remember that derivative of sinx is cosx, derivative of cosx is –sinx , it is a general mistake that the student confuse in the sign of the derivatives of sinx and cosx. It is also important that you know the derivatives of all the standard functions as they are very often used. We cannot directly find $\dfrac{dy}{dx}$ they have to differentiate with respect to $\theta $ and at the end they have to substitute $\theta =\dfrac{\pi }{4}$ to get final answer.