# how would i answer thisWatch

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#1 Question 4

I worked out the first part, i got dy/dx = (ln(3)(3^x))+(-ln(3)(3^-x))

not sure how to do part b
Last edited by Nobodys2; 4 weeks ago
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4 weeks ago
#2
Substitute x = 1 (that's the x-coordinate of the given point) into your dy/dx to find the gradient at that point.
Then write the equation of the tangent in the form y = mx + c where m is the gradient you've just worked out.
Then find c by substituting in the x and y coordinates of the point.
Finally arrange in the form that the question requires.
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#3
(Original post by MarkFromWales)
Substitute x = 1 (that's the x-coordinate of the given point) into your dy/dx to find the gradient at that point.
Then write the equation of the tangent in the form y = mx + c where m is the gradient you've just worked out.
Then find c by substituting in the x and y coordinates of the point.
Finally arrange in the form that the question requires.
did that, my calculator gives exact form and im not sure how to rearrange into that form the question wants
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4 weeks ago
#4
What have you got so far?
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#5
(Original post by NotNotBatman)
What have you got so far?
worked out the derivative for part a, dy/dx = (ln(3)(3^x))+(-ln(3)(3^-x))
substitued x=1 and got a bunch of numbers
0
4 weeks ago
#6
(Original post by Nobodys2)
worked out the derivative for part a, dy/dx = (ln(3)(3^x))+(-ln(3)(3^-x))
substitued x=1 and got a bunch of numbers
So you have the gradient of the tangent at x=1. Substitute and the numerical value of the gradient into the form Where and is the value of the gradient at that point and rearrange.
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#7
yeah i do that, just dont get the form that the question wants
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4 weeks ago
#8
(Original post by Nobodys2)
yeah i do that, just dont get the form that the question wants
What do you get?
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#9
(Original post by NotNotBatman)
What do you get?
im pretty close, cant figure out the last step EDIT: never mind i got, thanks for help
1
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