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#1
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)

A
0
6 years ago
#2
(Original post by 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)

A
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)
0
6 years ago
#3
(Original post by 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)

A

Chain rule means you would need to differentiate the -x giving -1 hence the MINUS
0
6 years ago
#4
(Original post by 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
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.
0
#5
(Original post by 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 TenOfThem)
Chain rule means you would need to differentiate the -x giving -1 hence the MINUS
(Original post by 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)
Thanks all! Oh how silly I've been-

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
0
6 years ago
#6
(Original post by 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?
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
0
#7
(Original post by 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
cheers! i get it now!
0
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