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Exercise

ddx(cot⁡(x)sec⁡(x)+x2)\frac{d}{dx}\left(\frac{\cot\left(x\right)}{\sec\left(x\right)+x^2}\right)

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Final answer to the exercise

(−sec⁡(x)−x2)csc⁡(x)2+(−sec⁡(x)tan⁡(x)−2x)cot⁡(x)(sec⁡(x)+x2)2\frac{\left(-\sec\left(x\right)-x^2\right)\csc\left(x\right)^2+\left(-\sec\left(x\right)\tan\left(x\right)-2x\right)\cot\left(x\right)}{\left(\sec\left(x\right)+x^2\right)^2}

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ddx (cot(x)sec(x)+x2 )
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