Question

find dy/dx y=5x^4 + 3x^3/2 +6x

Answer 1

Answer:

hope you got your answer

Answer 2

Answer:

$\frac{dy}{dx}= 20x^{3} +\frac{9}{2}\sqrt{x} + 6$

Explanation:

Given:

$y=5x^{4} + 3x^{ \frac{3}{2} } +6x$

Find:

$\frac{dy}{dx} = \ ?$

Solution:

$\frac{dy}{dx} = \frac{d}{dx}(5x^{4} + 3x^{\frac{3}{2} } + 6x)$

⇒ $\frac{dy}{dx} = \frac{d}{dx}(5x^{4}) + \frac{d}{dx}(3x^{\frac{3}{2} }) + \frac{d}{dx}(6x)$

⇒ $\frac{dy}{dx}= 5 \times \frac{d}{dx}(x^{4}) + 3 \times \frac{d}{dx}(x^{\frac{3}{2} }) + 6 \times \frac{d}{dx}(x)$

⇒ $\frac{dy}{dx}= 5 \times 4x^{(4-1)} + 3 \times \frac{3}{2}x^{(\frac{3}{2} -1)} + 6 \times 1x^{(1-1)}$

⇒ $\frac{dy}{dx}= 5 \times 4x^{3} + 3 \times \frac{3}{2}x^{\frac{1}{2}} + 6 \times 1x^{0}$

⇒ $\frac{dy}{dx}= 20x^{3} +\frac{9}{2}x^{\frac{1}{2}} + 6 x^{0}$

⇒ $\frac{dy}{dx}= 20x^{3} +\frac{9}{2}\sqrt{x} + 6$

Important derivatives used:

• $\frac{d}{dx}(x^{n}) = nx^{n-1}$     where n is a constant

• $\frac{d}{dx}(kx) = k \ \frac{d}{dx}(x)$   where k is a constant

• $\frac{d}{dx}(x) = 1$

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