Derivate (cos(3x^2). (5x^3 -1)^1/3 +sin 4x^3)^4 \: \: \: \: find \: first \: derivative \\ ( cos(3x {}^(2) ) * ( \sqrt[3]{5x {}^(3) - 1} ) + \sin(4x {}^(3) ) {}^(4) ​

Question

Derivate (cos(3x^2). (5x^3 -1)^1/3 +sin 4x^3)^4


`\: \: \: \: find \: first \: derivative \\ ( cos(3x {}^(2) ) * ( \sqrt[3]{5x {}^(3) - 1} ) + \sin(4x {}^(3) ) {}^(4)`
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Answer 1

Explanation:

`(d)/(dx) [cos(3x^2) \sqrt[3]{5x^3 -1} +sin(4x^3)]^4\\\\=4[cos(3x^2) \sqrt[3]{5x^3 -1} +sin(4x^3)]^3\; (d)/(dx) [cos(3x^2) \sqrt[3]{5x^3 -1} +sin(4x^3)] --- eq(1)`

Lets look at the derivative part:

`(d)/(dx) [cos(3x^2) \sqrt[3]{5x^3 -1} +sin(4x^3)] \\\\= (d)/(dx)[cos(3x^2) \sqrt[3]{5x^3 -1} ] + (d)/(dx)[sin(4x^3)]\\\\=cos(3x^2) (d)/(dx)[ \sqrt[3]{5x^3 -1} ] + \sqrt[3]{5x^3 -1}(d)/(dx)[ cos(3x^2) ] + cos(4x^3) (d)/(dx)[4x^3]\\\\=cos(3x^2) (1)/(3) (5x^3 -1)^{(1)/(3) -1} (d)/(dx)[5x^3 -1] + \sqrt[3]{5x^3 -1} (-sin(3x^2))(d)/(dx)[ 3x^2] + cos(4x^3)[(4)(3)x^2]`

`=\frac{cos(3x^2) 5(3)x^2}{3(5x^3 - 1)^{(2)/(3) }} -\sqrt[3]{5x^3 -1}\; sin(3x^2) (3)(2)x + 12x^2 cos(4x^3)\\\\=\frac{5x^2cos(3x^2) }{(5x^3 - 1)^{(2)/(3) }} -6x\sqrt[3]{5x^3 -1}\; sin(3x^2) + 12x^2 cos(4x^3)`

Substituting in eq(1), we have:

`(d)/(dx) [cos(3x^2) \sqrt[3]{5x^3 -1} +sin(4x^3)]^4\\\\=4[cos(3x^2) \sqrt[3]{5x^3 -1} +sin(4x^3)]^3\; [\frac{5x^2cos(3x^2) }{(5x^3 - 1)^{(2)/(3) }} -6x\sqrt[3]{5x^3 -1}\; sin(3x^2) + 12x^2 cos(4x^3)]`