Derivatives

[dave turbo]

Is this correct?

y = 4/(3*sqrt x) + 1/(3x^2)

y' = 17/3x^2

y' = (2)(17/3) x

y' = 11.3x

 

[HallsofIvy]

Is this correct?

y = 4/(3*sqrt x) + 1/(3x^2)

y' = 17/3x^2

y' = (2)(17/3) x

y' = 11.3x

No, it isn't! First, the second line, y'= 17/3x^2 is ambiguous. Do you mean (17/3)x^2 or 17/(3x^2)? Second, both that and third line, y'= (2)(17/3)x, start "y'= " but are obviously not the same. Finally, even if they were correct, (2)(17/3) is NOT equal to 11.3.

 

[Quaid]

y = 4/(3*sqrt x) + 1/(3x^2)

We can first factor out the constants.

\(\displaystyle y = \dfrac{4}{3} \cdot \dfrac{1}{\sqrt{x}} + \dfrac{1}{3} \cdot \dfrac{1}{x^{2}}\)

From precalculus, do you remember how to rewrite \(\displaystyle \dfrac{1}{\sqrt{x}}\) and \(\displaystyle \dfrac{1}{x^{2}}\) using exponential notation?

 

[dave turbo]

We can first factor out the constants.

\(\displaystyle y = \dfrac{4}{3} \cdot \dfrac{1}{\sqrt{x}} + \dfrac{1}{3} \cdot \dfrac{1}{x^{2}}\)

From precalculus, do you remember how to rewrite \(\displaystyle \dfrac{1}{\sqrt{x}}\) and \(\displaystyle \dfrac{1}{x^{2}}\) using exponential notation?

ok so to do that

4/3 * 1/(sqrtx) + 1/3 * 1/(x^2)

1.33*(x^1/2) + 3*x^2

0.665(x^-1/2) + 6x

?

 

[Deleted member 4993]

ok so to do that

4/3 * 1/(sqrtx) + 1/3 * 1/(x^2)

1.33*(x^1/2) + 3*x^2

0.665(x^-1/2) + 6x

?

Above is incorrect.

f(x) = 4/3 * 1/(sqrtx) + 1/3 * 1/(x^2)

f(x) = 4/3 * x^(-1/2) + 1/3 * x^(-2)

Now use

\(\displaystyle \frac{d}{dx}{x^n} \ = \ n * x^{n-1}\)

then

f'(x) = 4/3 * (-1/2) * x^(-1/2 - 1) + 1/3 * (-2) * x^(-2 - 1)

simplify to final answer ....

 

[Quaid]

4/3 * 1/(sqrtx) + 1/3 * 1/(x^2)

1.33*(x^1/2) + 3*x^2

It looks like you need to review the meaning of negative exponents. The values highlighted in red above need to be negative, as shown in Subhotosh's reply.

Why did you change the constant 1/3 to 3? Is that a typo?

Also, your instructor may expect an exact answer, for this exercise. If so, you're not allowed to replace the constant 4/3 with a decimal approximation. You should check, before changing the exercise.

Questions? :cool: