Derivatives of Inverse Trigonometric Functions

Suggested Prerequesites:

Inverse Trigonometric functions, Derivatives of Trigonometric functions, Implicit differentiation, The Chain rule

Inverse trigonometric functions aren't used very frequently. Why, then so we care about their derivatives? One reason is simply that we'd like to be able to differentiate any function that's thrown at us. The other reason is that they'll be more useful later on, when studying integration.

We'll find the derivative of sin^-1(x) and tan^-1(x) by defining them implicitly and then differentiating. Let's start with sin:

y &sp;=&sp; sin^-1^(x)

$sin y &sp;=&sp; x$

$D_x_(sin y) &sp;=&sp; D_x_(x)$

$(cos y) y' &sp;=&sp; 1$

$y' &sp;=&sp; 1 cos (y) &sp;=&sp; 1 1 - sin^2^y &sp;=&sp; 1 1 - x^2^$

Our result can be restated as:

d dx sin^-1^(x) &sp;=&sp; 1 1  -  x^2^

We'll apply a similiar method to tangent:

y &sp;=&sp; tan^-1^(x)

$tan (y) &sp;=&sp; x$

$D_x_tan (y) &sp;=&sp; D_x_(x)$

$(sec^2^y) y' &sp;=&sp; 1$

$y' &sp;=&sp; 1 sec^2^(y) &sp;=&sp; 1 1 + tan^2^(y) &sp;=&sp; 1 1 + x^2^$

To restate this result:

d dx tan^-1^(x) &sp;=&sp; 1 1   +  x^2^

Now, the derivatives of the other inverse trigonometric functions can also be defined, but they turn out to not be useful. So, we'll not bother to do so here.

Some examples:

$d dx sin^-1^ (3x) &sp;=&sp; (1 1 - (3x)^2^) (3)$
$d dx tan^-1^x - 1 x +
1 &sp;=&sp; (1 1 + (x - 1 x + 1)^2^) ((x+1) - (x-1) (x+1)^2^) &sp;=&sp; 2 (x+1)^2^ + (x-1)^2^$

Exercises:

$y &sp;=&sp; x(sin^-1^x)^2^$
$y &sp;=&sp; tan^-1^(cos x)$
Solutions to the exercises | Back to the Calculus page | Back to the World Web Math top page

jjnichol@mit.edu