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TS EAMCET · Maths · Differentiation

If xexy=y+sin2x, then dydx at x=0 is

  1. A 0
  2. B -1
  3. C -2
  4. D 1
Verified Solution

Answer & Solution

Correct Answer

(D) 1

Step-by-step Solution

Detailed explanation

xexy=y+sin2x, so at x=0, 0·e0·y=y+sin20⇒y=0 ...(1) Now, differentiating both sides i.e. dxexydx=dydx+dsin2xdx ⇒xexyxdydx+y+exy=dydx+2sinxcosx (Using chain rule i.e. fgx'=f'gx·g'x and product rule i.e. u·v'=uv'+u'v) Putting x=0,…