Question

Calculus early transcendental functions. Find the derivative of the function.

Answer 1

We are given the following function:

$y=\frac{10}{\sqrt[3]{x}}-2x+3$

We are asked to determine the derivative.

$\frac{dy}{dx}=\frac{d}{dx}(\frac{10}{\sqrt[3]{x}})-\frac{d}{dx}(2x)+\frac{d}{dx}(3)$

For the first derivative, we will use the following rule of exponents:

$\frac{a}{\sqrt[x]{b}}=a(b)^{-\frac{1}{x}}$

Applying the rule:

$\frac{dy}{dx}=\frac{d}{dx}(10(x)^{-\frac{1}{3}})-\frac{d}{dx}(2x)+\frac{d}{dx}(3)$

Now we will apply the following rule of derivatives:

$\frac{d}{dx}(x^n)=nx^{n-1}$

Applying the rule:

$\frac{dy}{dx}=10(-\frac{1}{3})(x)^{-\frac{4}{3}}-\frac{d}{dx}(2x)+\frac{d}{dx}(3)$

For the second derivative we will use the following rule:

$\frac{d}{dx}(ax)=a$

Applying the rule:

$\frac{dy}{dx}=10(-\frac{1}{3})(x)^{-\frac{4}{3}}-2+\frac{d}{dx}(3)$

For the third derivative we will use the fact that the derivative of a constant is zero, therefore:

$\frac{dy}{dx}=10(-\frac{1}{3})(x)^{-\frac{4}{3}}-2$

Now we solve the product:

$\frac{dy}{dx}=(-\frac{10}{3})(x)^{-\frac{4}{3}}-2$