Most of the time, to take the derivative of a function given by a formula y = f(x), we can apply differentiation functions (refer to the common derivatives table) along with the product, quotient, and chain rule. Sometimes though, it is not possible to solve and get an exact formula for y. For example, how would you find the derivative of y(x) given the following function?
In this case, we use implicit differentiation, so called because we cannot find y' explicitly as a function of only x. The best we can do is find a formula for y' as a function of both x and y. The resulting equation is called an implicit formula for y because we cannot solve for y directly.
To find y', we need to take the derivative of the entire equation with respect to x. Therefore, every time we encounter an expression involving functions of y, we need to use the chain rule since y itself is already a function of x. If we take the derivative of both sides with respect to x, we first apply the chain rule to y3 to get:
Important: We use the d/dx notation to emphasize that we are taking the derivative of the entire equation with respect to x, not y. That's why we have the extra factor of y' pop out when calculating above.
Similarly, we apply the chain rule and product rule with respect to x to sin(xy) to get:
On the right-hand side,
Plugging in the previous results, we get:
We couldn't solve for y, but we can solve for y'. First, distribute the cos(xy) term to both y and xy':
Then group the terms with y' to the left:
Now factor out the y' term:
And isolate the y' term:
Other times, it is not possible to find the derivative y' by direct application of differentiation rules even when there is a formula for y. For example,
At first glance, we may think of using the power or exponential rule, but x appears in both the base and exponent, so we cannot use either of those rules.
Whenever there is a variable in an exponent, like the above case, it is often useful to take the natural log of that exponent because natural log rules and properties allow us to essentially bring the exponent down, making the expression easier to work with.
If we apply the above, we get ln(y) = xln(x), which is now an implicit formula for y. Then we can take the derivative of both sides with respect to x to get:
Since we know that y = xx, we substitute it into the above equation to get:
We can use implicit differentiation to derive derivatives of the inverse trigonometric functions such as tan-1(x) and sin-1(x).
Derivative of y = tan-1(x):
If we apply the tangent function to both sides of the equation, y = tan-1(x), we get:
Now, take the derivative of both sides:
Next, we plug the following trig identity, (sec(y))2 = (tan(y))2 + 1, into the above equation to get, then substitute x for tan(y) to get:
Derivative of y = sin-1(x)
If we apply the sine function to both sides of the equation y = sin-1(x), we get:
Now we take the derivative of both sides to get:
and rearrange the result to get:
Since , which can be arranged as , and ,
Note: An alarm bell may have been triggered in your head when we automatically assumed that . Not to worry! We knew that and not because if we took the negative square root, y' would be:
which is always negative. But this is wrong since the graph of y = sin-1(x) is:
which always has a positive slope. Since the slope of the tangent line of a function at any point is the function's derivative at that point, the derivative of sin-1(x) cannot be negative since the slope is always positive.