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f(x)=(e^cos(4x)(3-2x)^(1/2))/(x^2tan(2x))−6−5−4−3−2−10123456−3-2.5−2-1.5−1-0.500.511.522.53xy

Exercise

ddx((ecos4x3−2x)x2(tan2x))\frac{d}{dx}\left(\frac{\left(e^{cos4x}\sqrt{3-2x}\right)}{x^2\left(tan2x\right)}\right)

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Final answer to the exercise

(−4sin⁡(4x)+−13−2x+−2x−2sec⁡(2x)csc⁡(2x))3−2xecos⁡(4x)x2tan⁡(2x)\left(-4\sin\left(4x\right)+\frac{-1}{3-2x}+\frac{-2}{x}-2\sec\left(2x\right)\csc\left(2x\right)\right)\frac{\sqrt{3-2x}e^{\cos\left(4x\right)}}{x^2\tan\left(2x\right)}

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ddx ((ecos4x√3−2x)x2(tan2x) )
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