Question

Let $f(x)=(2x+7)^3$ and $g(x)=cos^2(4x)$. Find

(a)$f'(x)$

(b)$g'(x)$

Answer 1

$f'(x)=6(2x+7)^{2}$
$g'(x)=-8*cos(4x)*sin(4x)$

Answer 2

f(x) = (2x + 7)³
f(x) = (2x + 7)(2x + 7)(2x + 7)
f(x) = [2x(2x + 7) + 7(2x + 7)](2x + 7)
f(x) = [2x(2x) + 2x(7) + 7(2x) + 7(7)](2x + 7)
f(x) = (4x² + 14x + 14x + 49)(2x + 7)
f(x) = (4x² + 28x + 49)(2x + 7)
f(x) = 4x²(2x + 7) + 28x(2x + 7) + 49(2x + 7)
f(x) = 4x²(2x) + 4x²(7) + 28x(2x) + 28x(7) + 49(2x) + 49(7)
f(x) = 8x³ + 28x² + 56x² + 196x + 98x + 343
f(x) = 8x³ + 84x² + 294x + 343

$f'(x) = \frac{f(x + \delta x) - f(x)}{\delta x}$
$f'(x) = \frac{{([(8x^{3} + 24dx^{3} + 24d^{2}x^{3} + 3d^{3}x^{3})] + [84d^{2}x^{2} + 162dx^{2} + 84x^{2}] + [294x + 294dx] + 343}) - (8x^{3} + 84x^{2} + 294x + 343)}{dx}$
$f'(x) = \frac{(8x^{3} - 8x^{3}) + (84x^{2} - 84x^{2}) + (294x - 294x) + (343 - 343) + 24dx^{3} + 24d^{2}x^{3} + 3d^{3}x^{3} + 84d^{2}x^{2} + 162dx^{2} + 294dx}{dx}$
$f'(x) = \frac{24dx^{3} + 24d^{2}x^{3} + 3d^{3}x^{3} + 84d^{2}x^{2} + 162dx^{2} + 294dx}{dx}$
$f'(x) = 24x^{2} + 24dx^{2} + 3d^{2}x^{2} + 84dx + 162x + 294$
$f'(x) = 24x^{2} + 162x + 294$
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g(x) = cos²(4x)
g(x) = cos(4x)cos(4x)

$g'(x) = D\{cos(4x)\}cos(4x) + cos(4x)D\{cos(4x)\}$
$g'(x) = -4sin(4x)cos(4x) - 4sin(4x)cos(4x)$
$g'(x) = -8sin(4x)cos(4x)$
$g'(x) = -8[2sin(2x)cos(2x)][cos^{2}(2x) - sin^{2}(2x)]$
$g'(x) = -8[4sin(x)cos^{3}(x) - 4sin^{3}(x)cos(x)][cos^{4}(x) - 2sin^{2}(x)cos^{2}(x) + sin^{4}(x) - 4sin(x)cos(x)$
$g'(x) = -8[4sin(x)cos^{7}(x) - 8sin^{3}(x)cos^{5}(x) - 16sin^{2}(x)cos^{4}(x) + 4sin^{5}(x)cos^{3}(x) - 4sin^{3}(x)cos^{5}(x) + 8sin^{5}(x)cos^{3}(x) + 16sin^{4}(x)cos^{2}(x) + 4sin^{7}(x)cos^{3}(x)}]$
$g'(x) = -32sin(x)cos^{7}(x) + 64sin^{3}cos^{5}(x) + 128sin^{2}(x)cos^{4}(x) - 32sin^{5}(x)cos^{3}(x) + 32sin^{3}(x)cos^{5}(x) - 64sin^{5}(x)cos^{3}(x) - 128sin^{4}(x)cos^{2}(x) + 32sin^{7}cos^{3}(x)$