What is the derivative of #y = arctan(x - sqrt(1+x^2))#?

Question

Emma Aldridge · Accepted Answer

#(sqrt(1+x^2)-x)/(2sqrt(1+x^2)(x^2-xsqrt(1+x^2)+1))#

The derivative of #arctanx# is #d/dxarctanx=1/(1+x^2)#, so the chain rule tells us that when we have a function inside the arctangent function, #d/dxarctanu=1/(1+u^2)(du)/dx#.

Thus:

#d/dxarctan(x-sqrt(1+x^2))=1/(1+(x-sqrt(1+x^2))^2)d/dx(x-sqrt(1+x^2))#

Note that #(x-sqrt(1+x^2))^2=x^2-2xsqrt(1+x^2)+(1+x^2)#.

Also note that #d/dxsqrt(1+x^2)=d/dx(1+x^2)^(1/2)=1/2(1+x^2)^(-1/2)(2x)=x/sqrt(1+x^2)#.

#=1/(2x^2-2xsqrt(1+x^2)+2)(1-x/sqrt(1+x^2))#

#=1/2*1/(x^2-xsqrt(1+x^2)+1)((sqrt(1+x^2)-x)/sqrt(1+x^2))#

#=(sqrt(1+x^2)-x)/(2sqrt(1+x^2)(x^2-xsqrt(1+x^2)+1))#

Isabella Ackerman · Answer

undefined To find the derivative of $ y = \arctan(x - \sqrt{1+x^2}) $, we'll apply the chain rule.

Let $ u = x - \sqrt{1+x^2} $.

$ \frac{du}{dx} = 1 - \frac{1}{2\sqrt{1+x^2}} \cdot 2x = 1 - \frac{x}{\sqrt{1+x^2}} $.

Now, let $ y = \arctan(u) $.

$ \frac{dy}{du} = \frac{1}{1 + u^2} $.

Using the chain rule, we multiply $ \frac{dy}{du} $ and $ \frac{du}{dx} $:

$ \frac{dy}{dx} = \frac{1}{1 + (x - \sqrt{1+x^2})^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) $.

Simplify the expression:

$ \frac{dy}{dx} = \frac{1}{1 + x^2 - 2x\sqrt{1+x^2} + 1 + x^2} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) $.

$ = \frac{1}{2 + 2x^2 - 2x\sqrt{1+x^2}} \cdot \left(1 - \frac{x}{\sqrt{1+x^2}}\right) $.

Thus, the derivative of $ y $ with respect to $ x $ is:

$ \frac{dy}{dx} = \frac{1 - \frac{x}{\sqrt{1+x^2}}}{2 + 2x^2 - 2x\sqrt{1+x^2}} $.