Question

find the derivative of the following function.
y = 9x^{-\frac{5}{2}}+2x^{-\frac{1}{2}}+x^{3}-5
y = square

Explanation:

Step1: Apply power - rule to first term

The power - rule for differentiation is $\frac{d}{dx}(ax^n)=nax^{n - 1}$. For the term $9x^{-\frac{5}{2}}$, $n=-\frac{5}{2}$ and $a = 9$. So, $\frac{d}{dx}(9x^{-\frac{5}{2}})=9\times(-\frac{5}{2})x^{-\frac{5}{2}-1}=-\frac{45}{2}x^{-\frac{7}{2}}$.

Step2: Apply power - rule to second term

For the term $2x^{-\frac{1}{2}}$, $n = -\frac{1}{2}$ and $a = 2$. So, $\frac{d}{dx}(2x^{-\frac{1}{2}})=2\times(-\frac{1}{2})x^{-\frac{1}{2}-1}=-x^{-\frac{3}{2}}$.

Step3: Apply power - rule to third term

For the term $x^3$, $n = 3$ and $a = 1$. So, $\frac{d}{dx}(x^3)=3x^{3 - 1}=3x^2$.

Step4: Differentiate the constant term

The derivative of a constant $C$ is 0. For the constant term $-5$, $\frac{d}{dx}(-5)=0$.

Step5: Sum up the derivatives of each term

$y'=-\frac{45}{2}x^{-\frac{7}{2}}-x^{-\frac{3}{2}}+3x^2+0=-\frac{45}{2x^{\frac{7}{2}}}-\frac{1}{x^{\frac{3}{2}}}+3x^2$.

Answer:

$-\frac{45}{2x^{\frac{7}{2}}}-\frac{1}{x^{\frac{3}{2}}}+3x^2$