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Question Number 94774 by Hamida last updated on 20/May/20

Commented by peter frank last updated on 20/May/20

$$\mathrm{u}={({5\mathrm{x}}^2-2\mathrm{x})}^3$$$$\frac{\mathrm{du}}{\mathrm{dx}}=3{({5\mathrm{x}}^2-2\mathrm{x})}^2(10\mathrm{x}-2)$$$$\mathrm{t}=\sin {}^{2}4\mathrm{x}$$$$\frac{\mathrm{dt}}{\mathrm{dx}}=8\sin 4\mathrm{xcos}4\mathrm{x}=4\sin 8\mathrm{x}$$$$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}}.\frac{\mathrm{dt}}{\mathrm{dx}}$$$$$$

Answered by prakash jain last updated on 21/May/20

$$\frac{d}{dx}{(5x^3-2x)}^3{\sin }^2\left(4x\right)$$$$={(5x^3-2x)}^3\frac{d}{dx}{\sin }^{2}4x+{\sin }^{2}4x\frac{d}{dx}{(5x^3-2x)}^3$$$$={(5x^3-23)}^{2}2\sin 4x\cdot 4\cos 4x+$$$${\sin }^{2}4x\cdot 3{(5x^3-2x)}^2(15x^2-2)$$

Answered by mathmax by abdo last updated on 21/May/20

$$\frac{\mathrm{dy}}{\mathrm{dx}}={\mathrm{y}}^{{}^{\prime }}\left(\mathrm{x}\right)=3(10\mathrm{x}-2){({5\mathrm{x}}^2-2\mathrm{x})}^2{\sin }^2\left(4\mathrm{x}\right)$$$$+8\cos \left(4\mathrm{x}\right)\sin \left(4\mathrm{x}\right){({5\mathrm{x}}^2-2\mathrm{x})}^3$$$$={({5\mathrm{x}}^2-2\mathrm{x})}^2\sin \left(4\mathrm{x}\right)\{3\sin \left(4\mathrm{x}\right)(10\mathrm{x}-2)+8\cos \left(4\mathrm{x}\right){({5\mathrm{x}}^2-2\mathrm{x})}^2\}$$

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