Question

Differentiation of (x^3 + 2x - 3)^4.(x + cosx)^3

Answer 1

Answer:

The derivative of cos3 (x) is -3cos2 (x) sin(x).

Answer 2

Answer:

Required derivative is $(x + cosx)^{2} \times ( {x}^{3} + 2x - 3)^{3}[4(x + cosx)(3 {x }^{2} + 2) + 3( {x}^{3} + 2x - 3)(1 - sinx)]$

Step-by-step explanation:

Given,

$( {x}^{3} + 2x - 3)^{4}(x + cosx)^{3}$

Let,

$y = ( {x}^{3} + 2x - 3)^{4}(x + cosx)^{3}$

We are differentiating both side with respect to x.

$\frac{dy}{dx} = \frac{d}{dx} [( {x}^{3} + 2x - 3)^{4}(x + cosx)^{3}]$

$= \frac{d}{dx} [( {x}^{3} + 2x - 3)^{4}](x + cosx)^{3} + ( {x}^{3} + 2x - 3)^{4} \frac{d}{dx}[ (x + cosx)^{3} ]$

$= (x + cosx)^{3} \times 4( {x}^{3} + 2x - 3)^{3} \frac{d}{dx} ( {x}^{3} + 2x - 3) +( {x}^{3} + 2x - 3)^{4} \times 3(x + cosx)^{2} \times \frac{d}{dx} (x + cosx)$

$= (x + cosx)^{3} \times 4( {x}^{3} + 2x - 3)^{3} ( 3{x}^{2} + 2 ) +( {x}^{3} + 2x - 3)^{4} 3(x + cosx)^{2} \times (1 - sinx)$

$= (x + cosx)^{2} \times ( {x}^{3} + 2x - 3)^{3}[4(x + cosx)(3 {x }^{2} + 2) + 3( {x}^{3} + 2x - 3)(1 - sinx)]$

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