# maths differentiation question help

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#1
A curve has the equation y=ln3x - e^-2x
Show that the equation of the tangent at the point with an x-coordinate of 1 is y=((e^2+2)/e^2)x - ((e^2+3)/e^2)+ln3

Please help I have no idea how to do this question. Help is much appreciated thank you
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8 months ago
#2
(Original post by beautiful_tae)
A curve has the equation y=ln3x - e^-2x
Show that the equation of the tangent at the point with an x-coordinate of 1 is y=((e^2+2)/e^2)x - ((e^2+3)/e^2)+ln3

Please help I have no idea how to do this question. Help is much appreciated thank you
It's always the same process to find the equation of a tangent to a curve at a point, you need the coordinates of the point and the gradient of the tangent to the curve at that point, which you find by differentiating. (I hope this is obvious, but you do say you have "no idea" how to do the question, so I thought I'd check.)

Assuming you're OK with that, it should be easy to find the y-coordinate to go with the x-coordinate, so I'm assuming it's the differentiation that's the issue. Is it one of the pieces you have to differentiate that is being tricky? Do you know how to differentiate lx(3x)? How about e^-2x?
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#3
(Original post by Pangol)
It's always the same process to find the equation of a tangent to a curve at a point, you need the coordinates of the point and the gradient of the tangent to the curve at that point, which you find by differentiating. (I hope this is obvious, but you do say you have "no idea" how to do the question, so I thought I'd check.)

Assuming you're OK with that, it should be easy to find the y-coordinate to go with the x-coordinate, so I'm assuming it's the differentiation that's the issue. Is it one of the pieces you have to differentiate that is being tricky? Do you know how to differentiate lx(3x)? How about e^-2x?
Yes i know to differentiate that but just seems to not getting the final right answer.

Differentiation: dy/dx= 1/x + 2e^-2x

then, i sub in x=1 into the dy/dx but i got 1.270670566

after that, i sub in the 1.270670566 into the equation of y=ln3x - e^-2x and got 1.259...

When i put them values into the format of y-y1=m(x-x1), i did not get the final answer as shown in the question so Idk where it went wrong.
0
8 months ago
#4
(Original post by beautiful_tae)
Yes i know to differentiate that but just seems to not getting the final right answer.

Differentiation: dy/dx= 1/x + 2e^-2x

then, i sub in x=1 into the dy/dx but i got 1.270670566

after that, i sub in the 1.270670566 into the equation of y=ln3x - e^-2x and got 1.259...

When i put them values into the format of y-y1=m(x-x1), i did not get the final answer as shown in the question so Idk where it went wrong.
Why are you rounding your values when the answer you seek has nothing rounded at all??

Just work with exact values. Also, why are you subbing in the dy/dx value back into the y equation ??
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#5
(Original post by RDKGames)
Why are you rounding your values when the answer you seek has nothing rounded at all??

Just work with exact values. Also, why are you subbing in the dy/dx value back into the y equation ??
how do you find y-value?
0
8 months ago
#6
(Original post by beautiful_tae)
how do you find y-value?
Sub in x=1 into the equation for y.
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#7
(Original post by RDKGames)
Sub in x=1 into the equation for y.
Yup which is the same equation for y.

so like y=ln3 - e^-2
0
8 months ago
#8
(Original post by beautiful_tae)
Yes i know to differentiate that but just seems to not getting the final right answer.

Differentiation: dy/dx= 1/x + 2e^-2x

then, i sub in x=1 into the dy/dx but i got 1.270670566

after that, i sub in the 1.270670566 into the equation of y=ln3x - e^-2x and got 1.259...

When i put them values into the format of y-y1=m(x-x1), i did not get the final answer as shown in the question so Idk where it went wrong.
Well this is the problem. Don't just feed it into a calculator. When you put x = 1 into your differential, you get 1 + 2e^-2, or 1 + 2/e^2. It should be obvious from the form of the answer that you need to use exact values, and in this case your calculator cannot cope with that and gives you a numerical answer instead.
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