# MyMathLab Answers

## MyMathLab Answers Are The Ultimate Solutions In 2023 For Critical Math Questi**ons**

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**Does MyMathLab record?**

The MyMathLab software does not record anything that you do on the program.

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**Is MyMathLab free?**

No, MyMathLab is a paid service that requires you to purchase access.

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**How do I access MyMathLab results?**

Students can access their MyMathLab results by logging in to the website and clicking on “Results”.

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**How do I redo an assignment on MyMathLab?**

To redo an assignment on MyMathLab:

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– Click the “Assignments” tab and select any course that you would like to start with.

– Once a course opens up, click on it’s corresponding section or question number, depending on what type of assignment you want to redo.

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**Get 100% Correct Pearson Math Lab Answers – Guaranteed!**

**Q1. Find dy/dx,where y= ∛(x ^{2}+4)^{2}**

Answer:

dy/dx=d/du (∛(u^{2} ))×du/dx ,where u=(x^{2}+4)

= d/du(u^{(2/3)})×du/dx

=2/3 u^{(-1/3)}×du/dx

=2/3 (x^{2}+4)^{(-1/3)}×2x

=4/3 x/∛(x^{2}+4)

**Q2. If f(x) = x sinx,then find f'(x)**

Answer:

Apply the product rule

d/dx (f(x))g(x)) = d/dx(f(x)) g(x) + f(x)d/dx(g(x)) with f(x) = x and g(x) = sin(x):

f'(x)=d/dx (x sinx)=d/dx (x)*sinx+x*d/dx (sinx)

=sinx+x cosx

**Q3. Evaluate d/dx ((2 + 3 sinx)/(2 – 3 sinx ))**

Answer:

Apply the quotient rule

d/dx (f(x)/g(x)) = d/dx(f(x) g(x)) – f(x) d/dx(g(x)) / g^{2}(x)

d/dx ((2 + 3 sinx)/(2 – 3 sinx ))=((2-3sinx) d/dx (2+3sinx)-(2+3sinx)d/dx(2-3sinx))/(2-3sinx)^{2}

=(2-3sinx)(3cosx)-(2+3sinx)(-3cosx)/(2-3sinx)^{2}

= 3cosx(2-3sinx+2+3sinx)/(2-3sinx)^{2} =12cosx/(2-3sinx)^{2}

**Q4. Find dy/dx,where x=(θ+sinθ),y=(1-cosθ)**

Answer:

x=(θ+sinθ) y=(1-cosθ)

dx/dθ=d/dθ (θ+sinθ)=1+cosθ dy/dθ=d/dθ (1+cosθ)=-sinθ

By chain rule of differentiation dy/dx=dy/dθ×dθ/dθ

dy/dx=-sinθ×1/(1+cosθ)=-sinθ/(1+cosθ)=-(2 sin(θ/2) cos(θ/2))/(2 cos^{2}(θ/2) )=-tan(θ/2)

**Q5. Find dy/dx,where x ^{(2/3)}+y^{(2/3)}=a^{(2/3)}**

Answer:

x^{(2/3)}+y^{(2/3)}=a^{(2/3)}

d/dx (x^{(2/3)} )+d/dx (y^{(2/3)} )=d/dx (a^{(2/3) })

2/3 x^{(-1/3)}+2/3 y^{(-1/3)}*dy/dx=0

dy/dx=(-2/3 x^{(-1/3)})/(2/3 y^{(-1/3) })=-x^{(-1/3)}/y^{(-1/3)} =-y^{(1/3)}/x^{(1/3)} = -∛(y/x)

**Q6. Evaluate ∫(x ^{2}+x+1)/(x√x) dx**

Answer:

∫(x^{2}+x+1)/(x√x) dx=∫x^{2}/(x√x) dx+∫x/(x√x) dx+∫1/(x√x) dx

=∫√x dx+∫1/√x dx+∫1/(x√x) dx

=∫x^{(1/2)} dx+∫x^{(-1/2)} dx+∫x^{(-3/2)} dx

=2/3 x^{(3/2)}+2x^{(1/2)}-2x^{(-1/2)}+C

= 2/3 x√x+2√x-2/√x+C

Where C is the indefinite integral constant.

**Frequently Asked Questions**

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