Lesson 2.6: Derivatives of Inverse Trigonometric Functions

One of the remaining classes of functions we have to find derivatives for is the inverse trigonometric functions. Let’s begin with trying to differentiate y=asin x. We can equivalently write x=sin y. From this, we can draw the diagram shown to the right: y is an angle of a right triangle with hypotenuse 1 and sides x and sqrt(1-x^2).

Triangle diagram for differentiating y=asin x

Now we can implicitly differentiate the equation x=sin y: d/dx(x)=1 and d/dy(sin y)=cos y, so d/dx(sin y)=cos y (dy/dx). Therefore our derivative is 1=(dy/dx)cos y. From our diagram we get that cos y=sqrt(1-x^2), so our derivative is 1=(dy/dx)sqrt(1-x^2). Therefore dy/dx=1/sqrt(1-x^2). Since we started with y=asin x, we now know that d/dx(asin x)=1/sqrt(1-x^2).

Triangle diagram for differentiating y=acos x

We can find d/dx (acos x) similarly, using the diagram to the right. Given y=acos x we write x=cos y. Now d/dx (x)=1 and d/dx (cos y)=-(dy/dx)sin y. Therefore the derivative is 1=-(dy/dx)sin y = -(dy/dx)sqrt(1-x^2). We solve for dy/dx = -1/sqrt(1-x^2).

Triangle diagram for differentiating y=atan x

The process for finding d/dx (atan x) is analogous: x=tan y gives a derivative of 1=(dy/dx)sec^2 y. Now the diagram in question, shown to the right, tells us that sec y=sqrt(x^2+1), so sec^2 y=x^2+1. Our derivative is 1=(dy/dx)(x^2+1), so d/dx (atan x)=1/(x^2+1).

By analogous procedures, we can show that d/dx (asec x)=1/(x sqrt(x^2+1)), d/dx (acsc x)=-1/(x sqrt(x^2+1)), and d/dx (acot x)=-1/(x^2+1).

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