4

Let $f$ be a differentiable function such that $f(1)=1$ and the slope of the tangent line to the curve $y=f{[x*f(x*y)]^2 }$ at the point $A(1,1)$ is $3$.

Find all possible values of $f'(1)$ .

my solution $1=x^2*(f(x))^2 (f(x))^2=1/x^2 ((f(x))^2)'=(1/x^2)' 2*f(x)*(f(x))'=-2/x^3 x=1 => 2*f(1)*(f(1))'=-2 f(1)=-2/2*f(1)=-1$ Answer is $-1$. So is it the only answer? i doubt because i did not use the last condition

Conor O'Brien
  • 1,636
Orkhan
  • 43

1 Answers1

1

Applying the chain rule we have

$$y=f((xf(xy))^2)\implies y'=f'((xf(xy))^2)\cdot 2xf(xy)\cdot (f(xy)+xf'(xy)\cdot (y+xy')).$$ Since $y'(1)=3,y(1)=f(1)=1,$ we have

$$3=f'(1)\cdot 2\cdot (1+f'(1)\cdot (1+3)).$$ That is,

$$3=2f'(1)(1+4f'(1))=8f'(1)^2+2f'(1)$$ or

$$8f'(1)^2+2f'(1)-3=0.$$

Solve the quadratic equation and you will get two solutions.

mfl
  • 29,399