Question

Suppose f:[−1,1]→R is a function with the property that x2+f(x)2=1 and suppose f(3/5)=−4/5. Use implicit differentiation to compute f′(3/5).

Answer 1

What did you get?
Any ideas?

Answer 2

Since x2+f(x)2=1 holds on an open interval around x=3/5, we can differentiate both sides to discover something about f′(3/5).

ddx(x2+f(x)2)=2x+2f(x)f′(x).

On the other hand, ddx(1)=0.
Consequently, 2x+2f(x)f′(x)=0 for values of x around 3/5.

Plugging in x=3/5 yields 2⋅34+2f(34)f′(34)=0.

Solving, we find that f′(34)=−(3/5)/f(3/5).

But f(3/5)=−4/5.

Therefore f′(34)=−(3/5)/(−4/5)=3/4.

Answer 3

The idea at least is correct, but there seem to be typos throughout.
\[\frac{\mathrm d}{\mathrm dx}\left(x^2+f(x)^2\right)=\frac{\mathrm d}{\mathrm dx}1\implies 2x+f(x)f'(x)=0\]and so plugging \(x=\dfrac{3}{5}\) and using the fact that \(f\left(\dfrac{3}{5}\right)=-\dfrac{4}{5}\), you get
\[2\left(\frac{3}{5}\right)+\left(-\frac{4}{5}\right)f'\left(\frac{3}{5}\right)=0\implies f'\left(\frac{3}{5}\right)=\frac{\frac{6}{5}}{\frac{4}{5}}=\frac{3}{2}\]