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Hollaaaaa,

So I embarked upon this question in a C4 textbook and it's baffled me- I'd be so grateful if someone could help me work it out:

Find the equation of the tangent to the curve at the point (2 , 17/4)

Now it's not actually finding the equation I'm having problems with. It's just differentiating

I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

helphelphelp please

A

So I embarked upon this question in a C4 textbook and it's baffled me- I'd be so grateful if someone could help me work it out:

Find the equation of the tangent to the curve at the point (2 , 17/4)

Now it's not actually finding the equation I'm having problems with. It's just differentiating

I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

helphelphelp please

A

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#2

(Original post by

Hollaaaaa,

So I embarked upon this question in a C4 textbook and it's baffled me- I'd be so grateful if someone could help me work it out:

Find the equation of the tangent to the curve at the point (2 , 17/4)

Now it's not actually finding the equation I'm having problems with. It's just differentiating

I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

helphelphelp please

A

**Aso**)Hollaaaaa,

So I embarked upon this question in a C4 textbook and it's baffled me- I'd be so grateful if someone could help me work it out:

Find the equation of the tangent to the curve at the point (2 , 17/4)

Now it's not actually finding the equation I'm having problems with. It's just differentiating

I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

helphelphelp please

A

ln(y) = -x*ln(2)

(1/y)(dy/dx) = -ln(2)

dy/dx = (y)*(-ln2)

so, dy/dx = -(2ˆ-x)*ln(2)

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#3

**Aso**)

Hollaaaaa,

So I embarked upon this question in a C4 textbook and it's baffled me- I'd be so grateful if someone could help me work it out:

Find the equation of the tangent to the curve at the point (2 , 17/4)

Now it's not actually finding the equation I'm having problems with. It's just differentiating

I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

helphelphelp please

A

Chain rule means you would need to differentiate the -x giving -1 hence the MINUS

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#4

(Original post by

I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

A

**Aso**)I differentiated each term because if y=a^x then dy/dx = a^x ln(a)

right?

So using this i get, 2^x ln(2) + 2^-x ln(2)

but the correct differential according to the answer is 2^x ln(2) MINUS 2^-x ln(2)

A

Or if you wrote it as it would be clearer what's going on.

**Edit:**Oh, too slow.

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(Original post by

Well you have a "-x" in the exponent, rather than x. Perhaps let u=-x and use the chain rule for that part.

Or if you wrote it as it would be clearer what's going on.

**ghostwalker**)Well you have a "-x" in the exponent, rather than x. Perhaps let u=-x and use the chain rule for that part.

Or if you wrote it as it would be clearer what's going on.

**Edit:**Oh, too slow.
(Original post by

Chain rule means you would need to differentiate the -x giving -1 hence the MINUS

**TenOfThem**)Chain rule means you would need to differentiate the -x giving -1 hence the MINUS

(Original post by

you got the first part correct so i will differentiate only y=2ˆ-x .

ln(y) = -x*ln(2)

(1/y)(dy/dx) = -ln(2)

dy/dx = (y)*(-ln2)

so, dy/dx = -(2ˆ-x)*ln(2)

**good-guy**)you got the first part correct so i will differentiate only y=2ˆ-x .

ln(y) = -x*ln(2)

(1/y)(dy/dx) = -ln(2)

dy/dx = (y)*(-ln2)

so, dy/dx = -(2ˆ-x)*ln(2)

I didnt know you could just differentiate them seperately-

why is it that you can make y=2^x and y= 2^-x

because doesnt y = 2^x + 2^-x? so how can you just assume that each term is equal to y?

Sorry maybe im just being a bit

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#6

(Original post by

why is it that you can make y=2^x and y= 2^-x

because doesnt y = 2^x + 2^-x? so how can you just assume that each term is equal to y?

**Aso**)why is it that you can make y=2^x and y= 2^-x

because doesnt y = 2^x + 2^-x? so how can you just assume that each term is equal to y?

That is not what you do

If you had to differentiate y = 3x^3 - 5x^2 + 3x -2

wouldn't you just differentiate each term

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(Original post by

Neither term = y

That is not what you do

If you had to differentiate y = 3x^3 - 5x^2 + 3x -2

wouldn't you just differentiate each term

**TenOfThem**)Neither term = y

That is not what you do

If you had to differentiate y = 3x^3 - 5x^2 + 3x -2

wouldn't you just differentiate each term

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