Differentiation Question 318
Question: If $ y=a\sin x+b\cos x, $ then $ y^{2}+{{( \frac{dy}{dx} )}^{2}} $ is a
Options:
A) Function of x
B) Function of y
C) Function of x and y
D) Constant
Show Answer
Answer:
Correct Answer: D
Solution:
$ y=a\sin x+b\cos x $
Differentiating with respect to x, we get $ \frac{dy}{dx}=a\cos x-b\sin x $
Now $ {{( \frac{dy}{dx} )}^{2}}={{(a\cos x-b\sin x)}^{2}} $
$ =a^{2}{{\cos }^{2}}x+b^{2}{{\sin }^{2}}x-2ab\sin x\cos x $
and $ y^{2}={{(a\sin x+b\cos x)}^{2}} $
$ =a^{2}{{\sin }^{2}}x+b^{2}{{\cos }^{2}}x+2ab\sin x\cos x $
So, $ {{( \frac{dy}{dx} )}^{2}}+y^{2}=a^{2}({{\sin }^{2}}x+{{\cos }^{2}}x)+b^{2}({{\sin }^{2}}x+{{\cos }^{2}}x) $
Hence $ {{( \frac{dy}{dx} )}^{2}}+y^{2}=(a^{2}+b^{2}) $ = constant.
BETA
