Solved on Feb 22, 2024

Find the derivative of y with respect to x evaluated at x=1 for the following functions: (21) $y=\frac{2 x-1}{x+3}$ , (22) $y=\frac{5 x-2}{x^{2}+3}$ , (23) $y=\left(\frac{3 x+2}{x}\right)\left(x^{-5}+1\right)$ , (24) $y=\left(2 x^{7}-x^{2}\right)\left(\frac{x-1}{x+1}\right)$ .

STEP 1

Assumptions
1. We need to find the derivative of the given functions with respect to $x$ .
2. After finding the derivative, we will evaluate it at $x=1$ .
3. We will use the quotient rule for derivatives where necessary: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ .
4. We will use the product rule for derivatives where necessary: $(fg)' = f'g + fg'$ .
5. We will apply the power rule for derivatives where necessary: $(x^n)' = nx^{n-1}$ .

STEP 2

Find the derivative of $y=\frac{2x-1}{x+3}$ with respect to $x$ .
$y' = \left(\frac{2x-1}{x+3}\right)'$

STEP 3

Apply the quotient rule.
$y' = \frac{(2x-1)'(x+3) - (2x-1)(x+3)'}{(x+3)^2}$

STEP 4

Differentiate $2x-1$ and $x+3$ .
$(2x-1)' = 2$ $(x+3)' = 1$

STEP 5

Substitute the derivatives into the quotient rule.
$y' = \frac{2(x+3) - (2x-1)(1)}{(x+3)^2}$

STEP 6

Expand and simplify the numerator.
$y' = \frac{2x + 6 - 2x + 1}{(x+3)^2}$

STEP 7

Combine like terms in the numerator.
$y' = \frac{7}{(x+3)^2}$

STEP 8

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{(1+3)^2}$

STEP 9

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{16}$

STEP 10

Find the derivative of $y=\frac{5x-2}{x^2+3}$ with respect to $x$ .
$y' = \left(\frac{5x-2}{x^2+3}\right)'$

STEP 11

Apply the quotient rule.
$y' = \frac{(5x-2)'(x^2+3) - (5x-2)(x^2+3)'}{(x^2+3)^2}$

STEP 12

Differentiate $5x-2$ and $x^2+3$ .
$(5x-2)' = 5$ $(x^2+3)' = 2x$

STEP 13

Substitute the derivatives into the quotient rule.
$y' = \frac{5(x^2+3) - (5x-2)(2x)}{(x^2+3)^2}$

STEP 14

Expand and simplify the numerator.
$y' = \frac{5x^2 + 15 - 10x^2 + 4x}{(x^2+3)^2}$

STEP 15

Combine like terms in the numerator.
$y' = \frac{-5x^2 + 4x + 15}{(x^2+3)^2}$

STEP 16

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{-5(1)^2 + 4(1) + 15}{(1^2+3)^2}$

STEP 17

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{-5 + 4 + 15}{16}$

STEP 18

Simplify the numerator.
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{14}{16}$

STEP 19

Reduce the fraction.
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{8}$

STEP 20

Find the derivative of $y=\left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)$ with respect to $x$ .
$y' = \left(\left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)\right)'$

STEP 21

Apply the product rule.
$y' = \left(\frac{3x+2}{x}\right)'\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)'$

STEP 22

Apply the quotient rule to $\left(\frac{3x+2}{x}\right)'$ and the power rule to $\left(x^{-5}+1\right)'$ .
$y' = \frac{(3x+2)'x - (3x+2)x'}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(x^{-5}+1)'$

STEP 23

Differentiate $3x+2$ , $x$ , and $x^{-5}+1$ .
$(3x+2)' = 3$ $x' = 1$ $(x^{-5}+1)' = -5x^{-6}$

STEP 24

Substitute the derivatives into the product and quotient rules.
$y' = \frac{3x - (3x+2)}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(-5x^{-6})$

STEP 25

Expand and simplify the expressions.
$y' = \frac{3x - 3x - 2}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(-5x^{-6})$

STEP 26

Combine like terms and simplify the fractions.
$y' = \frac{-2}{x^2}\left(x^{-5}+1\right) - \frac{5(3x+2)}{x^7}$

STEP 27

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{-2}{1^2}\left(1^{-5}+1\right) - \frac{5(3(1)+2)}{1^7}$

STEP 28

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = -2(1+1) - 5(3+2)$

STEP 29

Simplify the expression.
$\left.\frac{dy}{dx}\right|_{x=1} = -2(2) - 5(5)$
$\left.\frac{dy}{dx}\right|_{x=1} = -4 - 25$
$\left.\frac{dy}{dx}\right|_{x=1} = -29$

STEP 30

Find the derivative of $y=\left(2x^7-x^2\right)\left(\frac{x-1}{x+1}\right)$ with respect to $x$ .
$y' = \left(\left(2x^7-x^2\right)\left(\frac{x-1}{x+1}\right)\right)'$

STEP 31

Apply the product rule.
$y' = (2x^7-x^2)'\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{x-1}{x+1}\right)'$

STEP 32

Apply the power rule to $(2x^7-x^2)'$ and the quotient rule to $\left(\frac{x-1}{x+1}\right)'$ .
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{(x-1)'(x+1) - (x-1)(x+1)'}{(x+1)^2}\right)$

STEP 33

Differentiate $x-1$ and $x+1$ .
$(x-1)' = 1$ $(x+1)' = 1$

STEP 34

Substitute the derivatives into the product and quotient rules.
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{1(x+1) - (x-1)(1)}{(x+1)^2}\right)$

STEP 35

Expand and simplify the expressions.
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{x+1 - x + 1}{(x+1)^2}\right)$

STEP 36

Combine like terms and simplify the fractions.
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{2}{(x+1)^2}\right)$

STEP 37

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = (14(1)^6-2(1))\left(\frac{1-1}{1+1}\right) + (2(1)^7-(1)^2)\left(\frac{2}{(1+1)^2}\right)$

STEP 38

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = (14-2)\left(\frac{0}{2}\right) + (2-1)\left(\frac{2}{4}\right)$

STEP 39

Simplify the expression.
$\left.\frac{dy}{dx}\right|_{x=1} = 0 + \frac{2}{4}$
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{2}$
The derivatives of the functions at $x=1$ are:
For 21. $\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{16}$
For 22. $\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{8}$
For 23. $\left.\frac{dy}{dx}\right|_{x=1} = -29$
For 24. $\left.\frac{dy}{dx}\right|_{x=1} = \frac{1}{2}$

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STEP 1

STEP 2

Find the derivative of y=2x−1x+3y=\frac{2x-1}{x+3}y=x+32x−1​ with respect to xxx.y′=(2x−1x+3)′y' = \left(\frac{2x-1}{x+3}\right)'y′=(x+32x−1​)′

STEP 3

Apply the quotient rule.y′=(2x−1)′(x+3)−(2x−1)(x+3)′(x+3)2y' = \frac{(2x-1)'(x+3) - (2x-1)(x+3)'}{(x+3)^2}y′=(x+3)2(2x−1)′(x+3)−(2x−1)(x+3)′​

STEP 4

Differentiate 2x−12x-12x−1 and x+3x+3x+3.(2x−1)′=2(2x-1)' = 2(2x−1)′=2 (x+3)′=1(x+3)' = 1(x+3)′=1

STEP 5

Substitute the derivatives into the quotient rule.y′=2(x+3)−(2x−1)(1)(x+3)2y' = \frac{2(x+3) - (2x-1)(1)}{(x+3)^2}y′=(x+3)22(x+3)−(2x−1)(1)​

STEP 6

Expand and simplify the numerator.y′=2x+6−2x+1(x+3)2y' = \frac{2x + 6 - 2x + 1}{(x+3)^2}y′=(x+3)22x+6−2x+1​

STEP 7

Combine like terms in the numerator.y′=7(x+3)2y' = \frac{7}{(x+3)^2}y′=(x+3)27​

STEP 8

Evaluate the derivative at x=1x=1x=1.dydx∣x=1=7(1+3)2\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{(1+3)^2}dxdy​∣∣​x=1​=(1+3)27​

STEP 9

Calculate the value of the derivative at x=1x=1x=1.dydx∣x=1=716\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{16}dxdy​∣∣​x=1​=167​

STEP 10

Find the derivative of y=5x−2x2+3y=\frac{5x-2}{x^2+3}y=x2+35x−2​ with respect to xxx.y′=(5x−2x2+3)′y' = \left(\frac{5x-2}{x^2+3}\right)'y′=(x2+35x−2​)′

STEP 11

Apply the quotient rule.y′=(5x−2)′(x2+3)−(5x−2)(x2+3)′(x2+3)2y' = \frac{(5x-2)'(x^2+3) - (5x-2)(x^2+3)'}{(x^2+3)^2}y′=(x2+3)2(5x−2)′(x2+3)−(5x−2)(x2+3)′​

STEP 12

Differentiate 5x−25x-25x−2 and x2+3x^2+3x2+3.(5x−2)′=5(5x-2)' = 5(5x−2)′=5 (x2+3)′=2x(x^2+3)' = 2x(x2+3)′=2x

STEP 13

Substitute the derivatives into the quotient rule.y′=5(x2+3)−(5x−2)(2x)(x2+3)2y' = \frac{5(x^2+3) - (5x-2)(2x)}{(x^2+3)^2}y′=(x2+3)25(x2+3)−(5x−2)(2x)​

STEP 14

Expand and simplify the numerator.y′=5x2+15−10x2+4x(x2+3)2y' = \frac{5x^2 + 15 - 10x^2 + 4x}{(x^2+3)^2}y′=(x2+3)25x2+15−10x2+4x​

STEP 15

Combine like terms in the numerator.y′=−5x2+4x+15(x2+3)2y' = \frac{-5x^2 + 4x + 15}{(x^2+3)^2}y′=(x2+3)2−5x2+4x+15​

STEP 16

Evaluate the derivative at x=1x=1x=1.dydx∣x=1=−5(1)2+4(1)+15(12+3)2\left.\frac{dy}{dx}\right|_{x=1} = \frac{-5(1)^2 + 4(1) + 15}{(1^2+3)^2}dxdy​∣∣​x=1​=(12+3)2−5(1)2+4(1)+15​

STEP 17

Calculate the value of the derivative at x=1x=1x=1.dydx∣x=1=−5+4+1516\left.\frac{dy}{dx}\right|_{x=1} = \frac{-5 + 4 + 15}{16}dxdy​∣∣​x=1​=16−5+4+15​

STEP 18

Simplify the numerator.dydx∣x=1=1416\left.\frac{dy}{dx}\right|_{x=1} = \frac{14}{16}dxdy​∣∣​x=1​=1614​

STEP 19

Reduce the fraction.dydx∣x=1=78\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{8}dxdy​∣∣​x=1​=87​

STEP 20

Find the derivative of y=(3x+2x)(x−5+1)y=\left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)y=(x3x+2​)(x−5+1) with respect to xxx.y′=((3x+2x)(x−5+1))′y' = \left(\left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)\right)'y′=((x3x+2​)(x−5+1))′

STEP 21

Apply the product rule.y′=(3x+2x)′(x−5+1)+(3x+2x)(x−5+1)′y' = \left(\frac{3x+2}{x}\right)'\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)'y′=(x3x+2​)′(x−5+1)+(x3x+2​)(x−5+1)′

STEP 22

STEP 23

Differentiate 3x+23x+23x+2, xxx, and x−5+1x^{-5}+1x−5+1.(3x+2)′=3(3x+2)' = 3(3x+2)′=3 x′=1x' = 1x′=1 (x−5+1)′=−5x−6(x^{-5}+1)' = -5x^{-6}(x−5+1)′=−5x−6

STEP 24

Substitute the derivatives into the product and quotient rules.y′=3x−(3x+2)x2(x−5+1)+(3x+2x)(−5x−6)y' = \frac{3x - (3x+2)}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(-5x^{-6})y′=x23x−(3x+2)​(x−5+1)+(x3x+2​)(−5x−6)

STEP 25

Expand and simplify the expressions.y′=3x−3x−2x2(x−5+1)+(3x+2x)(−5x−6)y' = \frac{3x - 3x - 2}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(-5x^{-6})y′=x23x−3x−2​(x−5+1)+(x3x+2​)(−5x−6)

STEP 26

Combine like terms and simplify the fractions.y′=−2x2(x−5+1)−5(3x+2)x7y' = \frac{-2}{x^2}\left(x^{-5}+1\right) - \frac{5(3x+2)}{x^7}y′=x2−2​(x−5+1)−x75(3x+2)​

STEP 27

Evaluate the derivative at x=1x=1x=1.dydx∣x=1=−212(1−5+1)−5(3(1)+2)17\left.\frac{dy}{dx}\right|_{x=1} = \frac{-2}{1^2}\left(1^{-5}+1\right) - \frac{5(3(1)+2)}{1^7}dxdy​∣∣​x=1​=12−2​(1−5+1)−175(3(1)+2)​

STEP 28

Calculate the value of the derivative at x=1x=1x=1.dydx∣x=1=−2(1+1)−5(3+2)\left.\frac{dy}{dx}\right|_{x=1} = -2(1+1) - 5(3+2)dxdy​∣∣​x=1​=−2(1+1)−5(3+2)

STEP 29

STEP 30

Find the derivative of y=(2x7−x2)(x−1x+1)y=\left(2x^7-x^2\right)\left(\frac{x-1}{x+1}\right)y=(2x7−x2)(x+1x−1​) with respect to xxx.y′=((2x7−x2)(x−1x+1))′y' = \left(\left(2x^7-x^2\right)\left(\frac{x-1}{x+1}\right)\right)'y′=((2x7−x2)(x+1x−1​))′

STEP 31

Apply the product rule.y′=(2x7−x2)′(x−1x+1)+(2x7−x2)(x−1x+1)′y' = (2x^7-x^2)'\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{x-1}{x+1}\right)'y′=(2x7−x2)′(x+1x−1​)+(2x7−x2)(x+1x−1​)′

STEP 32

STEP 33

Differentiate x−1x-1x−1 and x+1x+1x+1.(x−1)′=1(x-1)' = 1(x−1)′=1 (x+1)′=1(x+1)' = 1(x+1)′=1

STEP 34

Substitute the derivatives into the product and quotient rules.y′=(14x6−2x)(x−1x+1)+(2x7−x2)(1(x+1)−(x−1)(1)(x+1)2)y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{1(x+1) - (x-1)(1)}{(x+1)^2}\right)y′=(14x6−2x)(x+1x−1​)+(2x7−x2)((x+1)21(x+1)−(x−1)(1)​)

STEP 35

Expand and simplify the expressions.y′=(14x6−2x)(x−1x+1)+(2x7−x2)(x+1−x+1(x+1)2)y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{x+1 - x + 1}{(x+1)^2}\right)y′=(14x6−2x)(x+1x−1​)+(2x7−x2)((x+1)2x+1−x+1​)

STEP 36

Combine like terms and simplify the fractions.y′=(14x6−2x)(x−1x+1)+(2x7−x2)(2(x+1)2)y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{2}{(x+1)^2}\right)y′=(14x6−2x)(x+1x−1​)+(2x7−x2)((x+1)22​)

STEP 37

Evaluate the derivative at x=1x=1x=1.dydx∣x=1=(14(1)6−2(1))(1−11+1)+(2(1)7−(1)2)(2(1+1)2)\left.\frac{dy}{dx}\right|_{x=1} = (14(1)^6-2(1))\left(\frac{1-1}{1+1}\right) + (2(1)^7-(1)^2)\left(\frac{2}{(1+1)^2}\right)dxdy​∣∣​x=1​=(14(1)6−2(1))(1+11−1​)+(2(1)7−(1)2)((1+1)22​)

STEP 38

Calculate the value of the derivative at x=1x=1x=1.dydx∣x=1=(14−2)(02)+(2−1)(24)\left.\frac{dy}{dx}\right|_{x=1} = (14-2)\left(\frac{0}{2}\right) + (2-1)\left(\frac{2}{4}\right)dxdy​∣∣​x=1​=(14−2)(20​)+(2−1)(42​)

STEP 39

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Find the derivative of $y=\frac{2x-1}{x+3}$ with respect to $x$ .
$y' = \left(\frac{2x-1}{x+3}\right)'$

Apply the quotient rule.
$y' = \frac{(2x-1)'(x+3) - (2x-1)(x+3)'}{(x+3)^2}$

Differentiate $2x-1$ and $x+3$ .
$(2x-1)' = 2$ $(x+3)' = 1$

Substitute the derivatives into the quotient rule.
$y' = \frac{2(x+3) - (2x-1)(1)}{(x+3)^2}$

Expand and simplify the numerator.
$y' = \frac{2x + 6 - 2x + 1}{(x+3)^2}$

Combine like terms in the numerator.
$y' = \frac{7}{(x+3)^2}$

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{(1+3)^2}$

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{16}$

Find the derivative of $y=\frac{5x-2}{x^2+3}$ with respect to $x$ .
$y' = \left(\frac{5x-2}{x^2+3}\right)'$

Apply the quotient rule.
$y' = \frac{(5x-2)'(x^2+3) - (5x-2)(x^2+3)'}{(x^2+3)^2}$

Differentiate $5x-2$ and $x^2+3$ .
$(5x-2)' = 5$ $(x^2+3)' = 2x$

Substitute the derivatives into the quotient rule.
$y' = \frac{5(x^2+3) - (5x-2)(2x)}{(x^2+3)^2}$

Expand and simplify the numerator.
$y' = \frac{5x^2 + 15 - 10x^2 + 4x}{(x^2+3)^2}$

Combine like terms in the numerator.
$y' = \frac{-5x^2 + 4x + 15}{(x^2+3)^2}$

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{-5(1)^2 + 4(1) + 15}{(1^2+3)^2}$

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{-5 + 4 + 15}{16}$

Simplify the numerator.
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{14}{16}$

Reduce the fraction.
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{7}{8}$

Find the derivative of $y=\left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)$ with respect to $x$ .
$y' = \left(\left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)\right)'$

Apply the product rule.
$y' = \left(\frac{3x+2}{x}\right)'\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)\left(x^{-5}+1\right)'$

Differentiate $3x+2$ , $x$ , and $x^{-5}+1$ .
$(3x+2)' = 3$ $x' = 1$ $(x^{-5}+1)' = -5x^{-6}$

Substitute the derivatives into the product and quotient rules.
$y' = \frac{3x - (3x+2)}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(-5x^{-6})$

Expand and simplify the expressions.
$y' = \frac{3x - 3x - 2}{x^2}\left(x^{-5}+1\right) + \left(\frac{3x+2}{x}\right)(-5x^{-6})$

Combine like terms and simplify the fractions.
$y' = \frac{-2}{x^2}\left(x^{-5}+1\right) - \frac{5(3x+2)}{x^7}$

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = \frac{-2}{1^2}\left(1^{-5}+1\right) - \frac{5(3(1)+2)}{1^7}$

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = -2(1+1) - 5(3+2)$

Find the derivative of $y=\left(2x^7-x^2\right)\left(\frac{x-1}{x+1}\right)$ with respect to $x$ .
$y' = \left(\left(2x^7-x^2\right)\left(\frac{x-1}{x+1}\right)\right)'$

Apply the product rule.
$y' = (2x^7-x^2)'\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{x-1}{x+1}\right)'$

Differentiate $x-1$ and $x+1$ .
$(x-1)' = 1$ $(x+1)' = 1$

Substitute the derivatives into the product and quotient rules.
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{1(x+1) - (x-1)(1)}{(x+1)^2}\right)$

Expand and simplify the expressions.
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{x+1 - x + 1}{(x+1)^2}\right)$

Combine like terms and simplify the fractions.
$y' = (14x^6-2x)\left(\frac{x-1}{x+1}\right) + (2x^7-x^2)\left(\frac{2}{(x+1)^2}\right)$

Evaluate the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = (14(1)^6-2(1))\left(\frac{1-1}{1+1}\right) + (2(1)^7-(1)^2)\left(\frac{2}{(1+1)^2}\right)$

Calculate the value of the derivative at $x=1$ .
$\left.\frac{dy}{dx}\right|_{x=1} = (14-2)\left(\frac{0}{2}\right) + (2-1)\left(\frac{2}{4}\right)$