For explicit functions I can calculate the derivative at a certian point using the original function: $$\frac{f(1+0.1) – f(1)}{0.1}$$
And then use $\frac{d}{dx}f(1)$ to check if the function is correct. But what can I do for implicit funcions? how can I calculate the change then compare it with my derivative function
Edit: for example if I was asked to differentiate
$f(x)=x^2+\tanh(x)$ and if I am unsure about my answer I could type in on the calculator:
$$\frac{f(5+0.000001) – f(5)}{0.000001}$$ and then check my $\frac{d}{dx}f(5)$ they should be approximately equal.
My question is if I have an implicit function like $$xy^3=\tan(x+2y)-(x^2-1)$$ and after I differentiate it how can I check if it is correct
Best Answer
You can first calculate the real derivative for a pair of $(x,y)$ that satisfies the equation, then choose a $Δx$, then test if $\left(x+\Delta x,y+\Delta x\frac{dy}{dx}\right)$ approximately satisfies the equation.
For example, looks like $\left(0, -\pi/8\right)$ is a root of your equation. So you can choose $x' = 0.000001$ and find the corresponding $y'$ using the derivative:
$$\begin{align*} 3xy^2\frac{dy}{dx} + y^3 &= \sec^2(x+2y)\left(1+2\frac{dy}{dx}\right) -2x\\ 3xy^2\frac{dy}{dx} -2\sec^2(x+2y)\frac{dy}{dx} &= \sec^2(x+2y) -2x - y^3\\ \frac{dy}{dx} &= \frac{\sec^2(x+2y) -2x - y^3}{3xy^2 -2\sec^2(x+2y)}\\ \left.\frac{dy}{dx}\right|_{\left(0,-\frac\pi8\right)} &= \frac{2+\frac{\pi^3}{8^3}}{-4} \end{align*}$$
So let $$y' = -\frac\pi8 +\frac{2+\frac{\pi^3}{8^3}}{-4}\cdot0.000001 \approx -0.392699596839\ldots$$
$$LHS = x'\left(y'\right)^3 = -6.0559372465\ldots×10^{-8}$$ $$RHS = \tan\left(x'+2y'\right)-\left(x'\right)^2 + 1 = -6.056\ldots×10^{-8}$$
And hope it is close enough...