Differentiate \( x^{2}=\frac{1}{x^{2}} \) w.y, \( +x^{4} \)

To differentiate the equation \( x^2 = \frac{1}{x^2} \) with respect to \( x \), we first rewrite it as \( x^4 - 1 = 0 \), which indicates the function is defined by \( f(x) = x^4 - 1 \). Now, differentiate using the power rule: \[ \frac{d}{dx}(x^4) - \frac{d}{dx}(1) = 0 \] This results in: \[ 4x^3 = 0 \] Thus, \( x^3 = 0 \), leading to \( x = 0 \) as the critical point. However, differentiate \( \frac{1}{x^2} \) using the chain rule to confirm. The derivative is: \[ \frac{d}{dx}\left(x^2\right) = 2x \quad \text{and} \quad \frac{d}{dx}\left(\frac{1}{x^2}\right) = -\frac{2}{x^3} \] Setting both derivatives equal results in determining the points of interest further down the line! Additionally, remember that when you're working with implicit differentiation, consider checking whether the function is defined for all values of \( x \)—in this case, \( x=0 \) is a point where the original function breaks down. So be cautious! For future practice, try differentiating similar equations or explore implicit differentiation techniques—it’s quite illuminating and helps deepen your understanding of calculus!