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.

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

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

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

Applying the rule:

`(dy)/(dx)=(d)/(dx)(10(x)^{-(1)/(3)})-(d)/(dx)(2x)+(d)/(dx)(3)`

Now we will apply the following rule of derivatives:

`(d)/(dx)(x^n)=nx^(n-1)`

Applying the rule:

`(dy)/(dx)=10(-(1)/(3))(x)^{-(4)/(3)}-(d)/(dx)(2x)+(d)/(dx)(3)`

For the second derivative we will use the following rule:

`(d)/(dx)(ax)=a`

Applying the rule:

`(dy)/(dx)=10(-(1)/(3))(x)^{-(4)/(3)}-2+(d)/(dx)(3)`

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

`(dy)/(dx)=10(-(1)/(3))(x)^{-(4)/(3)}-2`

Now we solve the product:

`(dy)/(dx)=(-(10)/(3))(x)^{-(4)/(3)}-2`