Question

(ii) find $\frac{d^{5}y}{dx^{5}}$ for $y = \frac{1}{x^{3}}+7x^{4}+10x^{2}+3x + 1$.

Explanation:

Step1: Rewrite the function

Rewrite $\frac{1}{x^{3}}$ as $x^{- 3}$. So $y=x^{-3}+7x^{4}+10x^{2}+3x + 1$.

Step2: Recall the power - rule for differentiation

The power - rule states that if $y = x^{n}$, then $\frac{dy}{dx}=nx^{n - 1}$.

Step3: Differentiate the first term $x^{-3}$ five times

For $y_1=x^{-3}$, the first - derivative: $\frac{dy_1}{dx}=-3x^{-4}$.
The second - derivative: $\frac{d^{2}y_1}{dx^{2}}=(-3)(-4)x^{-5}=12x^{-5}$.
The third - derivative: $\frac{d^{3}y_1}{dx^{3}}=(12)(-5)x^{-6}=-60x^{-6}$.
The fourth - derivative: $\frac{d^{4}y_1}{dx^{4}}=(-60)(-6)x^{-7}=360x^{-7}$.
The fifth - derivative: $\frac{d^{5}y_1}{dx^{5}}=(360)(-7)x^{-8}=-2520x^{-8}$.

Step4: Differentiate the second term $7x^{4}$ five times

For $y_2 = 7x^{4}$, the first - derivative: $\frac{dy_2}{dx}=28x^{3}$.
The second - derivative: $\frac{d^{2}y_2}{dx^{2}}=84x^{2}$.
The third - derivative: $\frac{d^{3}y_2}{dx^{3}}=168x$.
The fourth - derivative: $\frac{d^{4}y_2}{dx^{4}}=168$.
The fifth - derivative: $\frac{d^{5}y_2}{dx^{5}}=0$.

Step5: Differentiate the third term $10x^{2}$ five times

For $y_3 = 10x^{2}$, the first - derivative: $\frac{dy_3}{dx}=20x$.
The second - derivative: $\frac{d^{2}y_3}{dx^{2}}=20$.
The third - derivative: $\frac{d^{3}y_3}{dx^{3}}=0$.
The fourth - derivative: $\frac{d^{4}y_3}{dx^{4}}=0$.
The fifth - derivative: $\frac{d^{5}y_3}{dx^{5}}=0$.

Step6: Differentiate the fourth term $3x$ five times

For $y_4 = 3x$, the first - derivative: $\frac{dy_4}{dx}=3$.
The second - derivative: $\frac{d^{2}y_4}{dx^{2}}=0$.
The third - derivative: $\frac{d^{3}y_4}{dx^{3}}=0$.
The fourth - derivative: $\frac{d^{4}y_4}{dx^{4}}=0$.
The fifth - derivative: $\frac{d^{5}y_4}{dx^{5}}=0$.

Step7: Differentiate the fifth term $1$ five times

For $y_5 = 1$, $\frac{dy_5}{dx}=0$, $\frac{d^{2}y_5}{dx^{2}}=0$, $\frac{d^{3}y_5}{dx^{3}}=0$, $\frac{d^{4}y_5}{dx^{4}}=0$, $\frac{d^{5}y_5}{dx^{5}}=0$.

Step8: Sum up the fifth - derivatives of all terms

$\frac{d^{5}y}{dx^{5}}=\frac{d^{5}y_1}{dx^{5}}+\frac{d^{5}y_2}{dx^{5}}+\frac{d^{5}y_3}{dx^{5}}+\frac{d^{5}y_4}{dx^{5}}+\frac{d^{5}y_5}{dx^{5}}=-2520x^{-8}+0 + 0+0 + 0=-\frac{2520}{x^{8}}$.

Answer:

$-\frac{2520}{x^{8}}$