Video Transcript
Find the derivative of the inverse
sine function with respect to 𝑥.
We’re differentiating the inverse
sin of 𝑥 with respect to 𝑥. So we’re going to begin by letting
𝑦 be equal to the inverse sin of 𝑥. Then we can say that 𝑥 must be
equal to sin of 𝑦. We’re going to differentiate both
sides of this equation with respect to 𝑥. So we say that d by d𝑥 of 𝑥 is
equal to d by d𝑥 of sin of 𝑦. Well, the derivative of 𝑥 with
respect to 𝑥 is quite straight forward; it’s one. But we’re going to need to use
implicit differentiation which is a special case of the chain rule to differentiate
sin 𝑦 with respect to 𝑥.
The derivative of sin 𝑦 with
respect to 𝑦 is cos 𝑦. So the derivative of sin 𝑦 with
respect to 𝑥 is cos 𝑦 times the derivative of 𝑦 with respect to 𝑥 which is just
d𝑦 by d𝑥. So we currently see that one is
equal to cos of 𝑦 times d𝑦 by d𝑥. We divide both sides of this
equation by cos 𝑦 to form an equation for the derivative. And we see that d𝑦 by d𝑥 is equal
to one over cos 𝑦. Now, we’ve got a bit of a
problem. We do want an expression for the
derivative in terms of 𝑥 not 𝑦.
And remember, we said that 𝑥 was
equal to sin of 𝑦. So we’ll use the identity cos
squared 𝜃 plus sin squared 𝜃 equals one. And I’ve replaced 𝜃 with 𝑦. We’ll subtract sin squared 𝑦 from
both sides of the equation. And then we’ll take the square root
of both sides. And we see that cos of 𝑦 is equal
to the positive and negative square root of one minus sin squared 𝑦. Remember, inverse sin is restricted
to the closed interval negative 𝜋 by two to 𝜋 by two.
By our definition, that means 𝑦
must be greater than or equal to 𝜋 two and less than or equal to 𝜋 by two which,
in turn, means cos of 𝑦 must be greater than or equal to zero and less than or
equal to one. And that’s because in the interval
𝑦 is greater than or equal to negative 𝜋 by two and less than or equal to 𝜋 by
two. The smallest value cos of 𝑦 we’ll
take is zero. And the greatest value is one. And what this means here is we’re
going to take the positive square root of one minus sin squared 𝑦 only.
We can now replace sin of 𝑦 with
𝑥. And we see that cos of 𝑦 is equal
to the square root of one minus 𝑥 squared. And, therefore, d𝑦 by d𝑥 equals
one over the square root of one minus 𝑥 squared. And we found the derivative of the
inverse sin of 𝑥. It’s one over the square root of
one minus 𝑥 squared for values of 𝑥 in the range 𝑥 is greater than negative one
and less than one.
