Find the derivative of tan^(-1)((1)/(8x+9)).

Question

Find the derivative of  tan^(-1)((1)/(8x+9)).

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Identify Functions: We need to find the derivative of the function $ y = \tan^{-1}\left(\frac{1}{8x+9}\right) $. To do this, we will use the chain rule, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.Derivative of Outer Function: First, let's identify the outer function and the inner function. The outer function is $ \tan^{-1}(u) $, where $ u $ is the inner function. The inner function is $ \frac{1}{8x+9} $.Derivative of Inner Function: The derivative of the outer function $ \tan^{-1}(u) $ with respect to $ u $ is $ \frac{1}{1+u^2} $. We will apply this after we find the derivative of the inner function.Apply Chain Rule: The derivative of the inner function $ \frac{1}{8x+9} $ with respect to $ x $ is found using the quotient rule or recognizing it as $ (8x+9)^{-1} $ and using the power rule. The derivative is $ -\frac{8}{(8x+9)^2} $.Simplify Expression: Now we apply the chain rule. The derivative of $ y $ with respect to $ x $ is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. This gives us: $ \frac{d}{dx} \left[ \tan^{-1}\left(\frac{1}{8x+9}\right) \right] = \frac{1}{1+\left(\frac{1}{8x+9}\right)^2} \cdot \left( -\frac{8}{(8x+9)^2} \right) $.Final Derivative: Simplify the expression. The $ (8x+9)^2 $ in the denominator of the derivative of the inner function cancels with the $ (8x+9)^2 $ in the $ 1+\left(\frac{1}{8x+9}\right)^2 $ term, leaving us with: $ \frac{d}{dx} \left[ \tan^{-1}\left(\frac{1}{8x+9}\right) \right] = -\frac{8}{1+(8x+9)^2} $.The final simplified form of the derivative is: $ \frac{d}{dx} \left[ \tan^{-1}\left(\frac{1}{8x+9}\right) \right] = -\frac{8}{1+64x^2+144x+81} $.

Find the derivative of $\tan^{-1}\left(\frac{1}{8x+9}\right)$ .

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Find the derivative of $\tan^{-1}\left(\frac{1}{8x+9}\right)$ .