uri2e4g

Answered

2022-07-09

Derivatives of a fraction function

An example of a fraction function is:

$y=\frac{-8x}{({x}^{2}+3{)}^{2}}$

The quotient rule says that if the function one wishes to differentiate, $f(x)$, can be written as:

$h(x)=\frac{f(x)}{g(x)}$

Then the derivative is (according to what I learned):

${h}^{\prime}(x)=\frac{{f}^{\prime}(x)g(x)-f(x){g}^{\prime}(x)}{(g(x){)}^{2}}$

Then I think the procedure is the following:

$\begin{array}{rl}{y}^{\prime}& =\frac{24({x}^{2}-1)}{(({x}^{2}+3{)}^{2}{)}^{2}}\\ & =\frac{24({x}^{2}-1)}{({x}^{2}+3{)}^{4}}\end{array}$

However, the solution is...

${y}^{\prime}=\frac{24({x}^{2}-1)}{({x}^{2}+3{)}^{3}}$

What are my mistakes?

What is the correct way to derivate fractions?

An example of a fraction function is:

$y=\frac{-8x}{({x}^{2}+3{)}^{2}}$

The quotient rule says that if the function one wishes to differentiate, $f(x)$, can be written as:

$h(x)=\frac{f(x)}{g(x)}$

Then the derivative is (according to what I learned):

${h}^{\prime}(x)=\frac{{f}^{\prime}(x)g(x)-f(x){g}^{\prime}(x)}{(g(x){)}^{2}}$

Then I think the procedure is the following:

$\begin{array}{rl}{y}^{\prime}& =\frac{24({x}^{2}-1)}{(({x}^{2}+3{)}^{2}{)}^{2}}\\ & =\frac{24({x}^{2}-1)}{({x}^{2}+3{)}^{4}}\end{array}$

However, the solution is...

${y}^{\prime}=\frac{24({x}^{2}-1)}{({x}^{2}+3{)}^{3}}$

What are my mistakes?

What is the correct way to derivate fractions?

Answer & Explanation

Jayvion Tyler

Expert

2022-07-10Added 23 answers

Your second step (after writing down the quotient rule) should be:

${y}^{\prime}=\frac{-8({x}^{2}+3{)}^{2}+8x\cdot 2\cdot 2x({x}^{2}+3)}{(({x}^{2}+3{)}^{2}{)}^{2}},$

and then an ${x}^{2}+3$ cancels off and gives you the correct answer:

$\frac{-8({x}^{2}+3{)}^{2}+8x\cdot 2\cdot 2x({x}^{2}+3)}{(({x}^{2}+3{)}^{2}{)}^{2}}=\frac{-8({x}^{2}+3)+32{x}^{2}}{({x}^{2}+3{)}^{3}}=\frac{24({x}^{2}-1)}{({x}^{2}+3{)}^{3}}.$

${y}^{\prime}=\frac{-8({x}^{2}+3{)}^{2}+8x\cdot 2\cdot 2x({x}^{2}+3)}{(({x}^{2}+3{)}^{2}{)}^{2}},$

and then an ${x}^{2}+3$ cancels off and gives you the correct answer:

$\frac{-8({x}^{2}+3{)}^{2}+8x\cdot 2\cdot 2x({x}^{2}+3)}{(({x}^{2}+3{)}^{2}{)}^{2}}=\frac{-8({x}^{2}+3)+32{x}^{2}}{({x}^{2}+3{)}^{3}}=\frac{24({x}^{2}-1)}{({x}^{2}+3{)}^{3}}.$

Rebecca Villa

Expert

2022-07-11Added 3 answers

Hint #1: If you write your fraction as

$HIGH/LOW,$

then the derivative for the quotient is "given" by the mnemonic

$\frac{LOW\cdot d(HIGH)-HIGH\cdot d(LOW)}{(LOW{)}^{2}}.$

Hint #2: $d(HIGH)=-8$

Hint #3: $d(LOW)=2({x}^{2}+3)(2x)=4x({x}^{2}+3)$

Can you take it from here?

$HIGH/LOW,$

then the derivative for the quotient is "given" by the mnemonic

$\frac{LOW\cdot d(HIGH)-HIGH\cdot d(LOW)}{(LOW{)}^{2}}.$

Hint #2: $d(HIGH)=-8$

Hint #3: $d(LOW)=2({x}^{2}+3)(2x)=4x({x}^{2}+3)$

Can you take it from here?

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