Question

examples of implicit differentiation?

Answer 1

Sure. Finding \(\ln(x)\).

Answer 2

Finding the derivative of it.

Answer 3

\[\begin{array}{ccc}
y&=&\ln(x)\\
e^y&=&x \\
e^yy'&=&1\\
xy'&=&1\\
y'&=& \frac{1}{x}
\end{array}
\]

Answer 4

Finding the derivative of \(\arcsin(x)\). \[
\begin{array}{ccc}
y&=&\arcsin(x)\\
\sin(y) &=& x\\
\cos(y)y' &=& 1\\
\cos(y) &=&\frac{1}{y'}\\
x^2+\left(\frac{1}{y'}\right)^2 &=& 1\\
\left(\frac{1}{y'}\right)^2 &=& 1-x^2 \\
\left|\frac{1}{y'}\right| &=& \sqrt{1-x^2} \\
\left|y'\right| &=& \frac{1}{\sqrt{1-x^2}} \\

\end{array}
\]

Answer 5

Technically everything is implicit differentiation, you're just not paying attention.

\[y=x^2\]
square root both sides, same as to 1/2 power.
\[y^{1/2}=x\]
take the derivative of both sides with respect to x.
\[\frac{ 1 }{ 2 }y^{-1/2}*\frac{ dy }{ dx }=1\]
Solve for dy/dx
\[\frac{ dy }{ dx }=2y^{1/2}\]
Ok looks weird, but just plug in y=x^2 and you get:
\[\frac{ dy }{ dx } = 2(x^2)^{1/2}=2x\]

Yeah, see, you were always doing the product and chain rule, you just had always done:\[\frac{ d }{dx}(y^1)=1*y^0*\frac{ dy }{ dx }\] without knowing that you took the 1 in the exponent, dropped it down front, subtracted 1, and then used the chain rule.

Similarly, with x:\[\frac{ d }{ dx }(x^1)=1*x^0*\frac{ dx }{ dx }=1\] since obviously for every amount of x you go x, so the change of x with respect to itself is exactly 1... clearly... lol.