Step by step solution to "How do I go about solving this derivative..."

Destinee Bryan

Answered question

2022-04-14

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

f (x) = 8 \tan^{- 1} (\frac{y}{x}) - \ln (\sqrt{x^{2} + y^{2}})

carlosegundoacyg

Beginner2022-04-15Added 10 answers

Since

f (x) = 8 \tan^{- 1} (\frac{y}{x}) - \ln (\sqrt{x^{2} + y^{2}})

since

8 \frac{d}{d x} \tan^{- 1} (x) = 8 \frac{1}{1 + x^{2}}

would

8 \frac{d}{d x} \tan^{- 1} (\frac{y}{x}) = 8 (\frac{1}{1 + {(\frac{y}{x})}^{2}})

\frac{d}{d x} (\tan^{- 1} (\frac{y}{x})) = \frac{8}{1 + {(\frac{y}{x})}^{2}} \cdot \frac{d}{d x} (\frac{y}{x})

Recall that

\frac{d}{d x} \tan^{- 1} (g (x)) = \frac{1}{1 + {(g (x))}^{2}} \cdot g^{'} (x)

Here,

g (x) = \frac{y}{x} \Rightarrow g^{'} (x) = \frac{- y}{x^{2}}

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

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

Now, when differentiating the second term, don't forget the chain rule for this term too.

\frac{d}{d x} (\frac{1}{2} \ln (x^{2} + y^{2})) = \frac{1}{2} \cdot \frac{1}{x^{2} + y^{2}} \cdot \frac{d}{d x} (x^{2} + y^{2})

And remember, y is being used as a constant.

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