How do I go about solving this derivative of inverse tangent?

Question

Okay so I have $$f(x)=8\tan^{-1}\left(\frac{y}{x}\right)-\ln \left(\sqrt{x^2+y^2}\right)$$ since $$8\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}(x)=8\frac{1}{1+x^2}$$would $$8\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}\left(\frac{y}{x}\right)=8\left(\frac{1}{1+\left(\frac{y}{x}\right)^2}\right)\text{ ??}$$ If not, how do I go about finding the derivative? Please if it's not right, show me step by step how to get the derivative so that I can fully grasp the process. Thanks in advance!! EDIT: SORRY I CHANGED THE NOTATION.

Answer 1

$$\dfrac{d}{dx}\left(\tan^{-1} \left(\frac yx\right)\right) = \frac 8{1+\left(\frac yx\right)^2}\cdot \dfrac{d}{dx}\left(\frac yx\right)$$

(You forgot to use the chain rule (see the last factor in the product on the right, above.) Recall that $$\frac{d}{dx} \tan^{-1}(g(x)) = \frac 1{1 + (g(x))^2} \cdot g'(x)$$ Here, $$g(x) = \frac yx \implies g'(x) = \frac {-y}{x^2}$$

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For the second term, use a nice property of logarithms:

$$\ln\sqrt{x^2 + y^2} = \ln(x^2 + y^2)^{1/2} = \frac 12 \ln(x^2 + y^2)$$

Now, when differentiating the second term, don't forget the chain rule for this term too. $$\frac d{dx}\left(\frac 12 \ln(x^2 + y^2)\right) = \frac 12\cdot \dfrac 1{x^2 + y^2}\cdot \frac d{dx}\left(x^2 + y^2\right)$$

And remember, $y$ is being used as a constant.

Can you take it from here?

Answer 2

If $y$ is a constant, treat it as a constant! For the first term you'll get $$ \frac{-\frac{y}{x^2}}{1+(\frac{y}{x})^2} $$

EDIT: for the second term use the property $\log x^a = a \log x $ and then apply the chain rule.