How to find d/dx ?

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How to find d/dx ?

d/dx {(1+ $x^{2} + x^{4}$ $x^2+x^4$ )^2x-3 }

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Answer 1 · 2018-02-17T05:45:21

$\frac{d}{d x}$ $d/dx$ = $(4 x - 6) {(x^{4} + x^{2} + 1)}^{2 x - 4} + (4 x^{3} + 2 x)$ $(4x-6)(x^4+x^2+1)^(2x-4)+(4x^3+2x)$

Explanation:

Applying the product rule;
Bring (2x-3) down, the power minus 1. Lastly, multiply by 2 because we have to differentiate 2x-3
Next we differentiate the value of $(x^{4} + x^{2} + 1)$ $(x^4+x^2+1)$ ,
So, $\frac{d}{d x}$ $d/dx$ = $(2 x - 3) {(x^{4} + x^{2} + 1)}^{(2 x - 3) - 1} (2) + (4 x^{3} + 2 x)$ $(2x-3)(x^4+x^2+1)^((2x-3)-1)(2)+(4x^3+2x)$
$\frac{d}{d x}$ $d/dx$ = $(4 x - 6) {(x^{4} + x^{2} + 1)}^{2 x - 4} + (4 x^{3} + 2 x)$ $(4x-6)(x^4+x^2+1)^(2x-4)+(4x^3+2x)$

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How to find d/dx ?

d/dx {(1+ $x^{2} + x^{4}$ $x^2+x^4$ )^2x-3 }

1 Answer

Explanation:

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How to find d/dx ?

d/dx {(1+x^2+x^4x2+x4x^2+x^4)^2x-3 }

1 Answer

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d/dx {(1+ $x^{2} + x^{4}$ $x^2+x^4$ )^2x-3 }