George S.

Unavailable

Physics (Masters) - Birmingham University

5.0

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5 completed lessons

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#### Ratings & Reviews

5from 2 customer reviews

Carole (Parent from Dudley)

January 31 2017

George is great. Patient, clear and concise.

Janviere (Student)

February 5 2017

#### Qualifications

MathsA-level (A2)A*
Further MathsA-level (A2)A*
PhysicsA-level (A2)A

#### General Availability

Pre 12pm12-5pmAfter 5pm
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#### Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£20 /hr
MathsA Level£20 /hr
PhysicsA Level£20 /hr
Further MathematicsGCSE£18 /hr
MathsGCSE£18 /hr
PhysicsGCSE£18 /hr
Maths13 Plus£18 /hr
Maths11 Plus£18 /hr

### How would I differentiate cos(2x)/x^1/2

So for this question you can use either the product rule or the quotient rule and I'll run through them both.

First the quotient rule:

The quotient rule says that if you have h(x)=f(x)/g(x)

Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2

So using f(x)=cos(2x) and g(x)=x^1/2

then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2

Plugging this into our formula gives us

h(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Always remember to simplify afterwards which gives us

(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Second the product rule:

What the product rule says is that if

h(x) = f(x)g(x)

then h'(x) = f(x)g'(x) + f'(x)g(x)

So if we say that h(x) = cos(2x)/x^1/2

Then we can say that f(x) = cos(2x) and g(x) = x^-1/2

Using the product rule we have:

f(x) = cos(2x)       f'(x) = -2sin(2x)

g(x) = x^-1/2      g'(x) = 1/2x^-3/2

So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)

So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2

Once again simplifying gives us

(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

So for this question you can use either the product rule or the quotient rule and I'll run through them both.

First the quotient rule:

The quotient rule says that if you have h(x)=f(x)/g(x)

Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2

So using f(x)=cos(2x) and g(x)=x^1/2

then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2

Plugging this into our formula gives us

h(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Always remember to simplify afterwards which gives us

(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Second the product rule:

What the product rule says is that if

h(x) = f(x)g(x)

then h'(x) = f(x)g'(x) + f'(x)g(x)

So if we say that h(x) = cos(2x)/x^1/2

Then we can say that f(x) = cos(2x) and g(x) = x^-1/2

Using the product rule we have:

f(x) = cos(2x)       f'(x) = -2sin(2x)

g(x) = x^-1/2      g'(x) = 1/2x^-3/2

So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)

So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2

Once again simplifying gives us

(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

2 years ago

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