How do you find the derivative of #tan(x − y) = x#?

Answer 1

#(dy)/(dx)=x^2/(1+x^2)#

I'm assuming you want to find #(dy)/(dx)#. For this we first need an expression for #y# in terms of #x#. We note that this problem has various solutions, since #tan(x)# is a periodic functions, #tan(x-y)=x# will have multiple solutions. However, since we know the period of the tangent function (#pi#), we can do the following: #x-y=tan^(-1)x+npi#, where #tan^(-1)# is the inverse function of the tangent giving values between #-pi/2# and #pi/2# and the factor #npi# has been added to account for the periodicity of the tangent.
This gives us #y=x-tan^(-1)x-npi#, therefore #(dy)/(dx)=1-d/(dx)tan^(-1)x#, note that the factor #npi# has disappeared. Now we need to find #d/(dx)tan^(-1)x#. This is quite tricky, but doable using the reverse function theorem.
Setting #u=tan^(-1)x#, we have #x=tanu=sinu/cosu#, so #(dx)/(du)=(cos^2u+sin^2u)/cos^2u=1/cos^2u#, using the quotient rule and some trigonometric identities. Using the inverse function theorem (which states that if #(dx)/(du)# is continuous and non-zero, we have #(du)/(dx)=1/((dx)/(du))#), we have #(du)/(dx)=cos^2u#. Now we need to express #cos^2u# in terms of x.
To do this, we use some trigonometry. Given a right triangle with sides #a,b,c# where #c# is the hypotenuse and #a,b# connected to the right angle. If #u# is the angle where side #c# intersects side #a#, we have #x=tanu=b/a#. With the symbols #a,b,c# in the equations we denote de length of these edges. #cosu=a/c# and using Pythagoras theorem, we find #c=sqrt(a^2+b^2)=asqrt(1+(b/a)^2)=asqrt(1+x^2)#. This gives #cosu=1/sqrt(1+x^2)#, so #(du)/(dx)=1/(1+x^2)#.
Since #u=tan^(-1)x#, we can substitute this into our equation for #(dy)/(dx)# and find #(dy)/(dx)=1-1/(1+x^2)=x^2/(1+x^2)#.
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Answer 2

To find the derivative of ( \tan(x - y) = x ), you'll use implicit differentiation. First, differentiate both sides of the equation with respect to ( x ), treating ( y ) as a function of ( x ). Then, solve for ( \frac{dy}{dx} ).

The derivative of ( \tan(x - y) ) with respect to ( x ) is ( \sec^2(x - y) \frac{d}{dx}(x - y) ).

Applying the chain rule to ( \frac{d}{dx}(x - y) ), we get ( 1 - \frac{dy}{dx} ).

Setting this equal to ( 1 ) (the derivative of ( x )), we have ( \sec^2(x - y)(1 - \frac{dy}{dx}) = 1 ).

Rearranging and solving for ( \frac{dy}{dx} ), we get ( \frac{dy}{dx} = 1 - \sec^2(x - y) ).

Therefore, the derivative of ( \tan(x - y) = x ) with respect to ( x ) is ( \frac{dy}{dx} = 1 - \sec^2(x - y) ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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