CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

Find $$ \displaystyle \frac{dy}{dx} $$ of $$\sin^2 y + \cos xy = k$$


Solution

We have, $$\sin^2 y + \cos xy = k$$
Differentiating both sides with respect to $$x$$, we obtain
$$\Rightarrow \displaystyle \frac{d}{dx} (\sin^2 y) + \frac{d}{dx} (\cos xy) = \frac{d(\pi)}{dx}=0$$ .....(1)
Using chain rule, we obtain
$$\displaystyle
\frac{d}{dx} (\sin^2 y) = 2 \sin y \frac{d}{dx} (\sin y) = 2 \sin y \cos
y \frac{dy}{dx}$$...... (2)
and 
$$\displaystyle
\frac{d}{dx} (\cos xy) = - \sin xy \frac{d}{dx} (xy) = - \sin xy \left[
y \frac{d}{dx} (x) + x \frac{dy}{dx} \right]$$
$$= - \sin xy \left[ y. 1 + x \dfrac{dy}{dx} \right] = -y \sin xy - x \sin xy \dfrac{dy}{dx}$$.....(3)
From (1), (2) and (3), we obtain
$$\displaystyle 2 \sin y \cos y \frac{dy}{dx} - y \sin xy - x \sin xy \frac{dy}{dx} = 0$$
$$\Rightarrow \displaystyle (2 \sin y \cos y - x \sin xy) \frac{dy}{dx} = y \sin xy$$
$$\Rightarrow \displaystyle (\sin 2 y - x \sin xy) \frac{dy}{dx} = y \sin xy$$
$$\therefore \displaystyle \frac{dy}{dx} = \frac{y \sin xy}{\sin 2 y - x \sin xy}$$

Mathematics
RS Agarwal
Standard XII

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
Same exercise questions
View More


similar_icon
People also searched for
View More



footer-image