QUESTION IMAGE
Question
rewriting a function before differentiating
in exercises 27–30, complete the table to find the derivative of the function.
original function\trewrite\tdifferentiate\tsimplify
- ( y = \frac{2}{7x^4} )\t\t\t
- ( y = \frac{8}{5x^{-5}} )\t\t\t
- ( y = \frac{6}{(5x)^3} )\t\t\t
- ( y = \frac{3}{(2x)^{-2}} )
Problem 27:
Step1: Rewrite the function
We can rewrite \( y = \frac{2}{7x^4} \) using the negative exponent rule \( \frac{1}{x^n}=x^{-n} \). So, \( y=\frac{2}{7}x^{-4} \) (since \( \frac{2}{7x^4}=\frac{2}{7}\cdot\frac{1}{x^4}=\frac{2}{7}x^{-4} \))
Step2: Differentiate using power rule
The power rule for differentiation is \( \frac{d}{dx}(x^n)=nx^{n - 1} \). Differentiating \( y=\frac{2}{7}x^{-4} \) with respect to \( x \), we get \( y^\prime=\frac{2}{7}\cdot(- 4)x^{-4-1} \)
Step3: Simplify the expression
Simplify \( \frac{2}{7}\cdot(-4)x^{-5}=-\frac{8}{7}x^{-5} \), and we can rewrite it with positive exponent as \( y^\prime =-\frac{8}{7x^5} \)
Step1: Rewrite the function
First, simplify \( \frac{8}{5x^{-5}} \). Using the rule \( \frac{1}{x^{-n}}=x^{n} \), we have \( \frac{8}{5}x^{5} \) (because \( \frac{8}{5x^{-5}}=\frac{8}{5}\cdot x^{5} \))
Step2: Differentiate using power rule
Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), differentiating \( y = \frac{8}{5}x^{5} \) gives \( y^\prime=\frac{8}{5}\cdot5x^{5 - 1} \)
Step3: Simplify the expression
Simplify \( \frac{8}{5}\cdot5x^{4}=8x^{4} \)
Step1: Rewrite the function
First, expand \( (5x)^3=5^3x^3 = 125x^3 \). So, \( y=\frac{6}{125x^3}=\frac{6}{125}x^{-3} \) (using \( \frac{1}{x^n}=x^{-n} \))
Step2: Differentiate using power rule
Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), differentiating \( y=\frac{6}{125}x^{-3} \) gives \( y^\prime=\frac{6}{125}\cdot(-3)x^{-3-1} \)
Step3: Simplify the expression
Simplify \( \frac{6}{125}\cdot(-3)x^{-4}=-\frac{18}{125}x^{-4}=-\frac{18}{125x^4} \)
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- Rewrite: \( y=\frac{2}{7}x^{-4} \)
- Differentiate: \( y^\prime=\frac{2}{7}\cdot(-4)x^{-5} \)
- Simplify: \( y^\prime =-\frac{8}{7x^5} \)