Find dy÷dx: y = (√x+1÷√x) sqr
Answers
Answered by
0
Answer:
y={(√x+1)/(√x)}^2
={1+1/(√x)}^2
={1+(1/x)+2/(√x}
dy/dx=-1/x^2+2(-1/2)(x^-3/2)
=-(1/x^2)-(1/x^(3/2))
Similar questions