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    Hey guys I'm just going through the uploaded question. I've managed part A, but on part B I'm unsure of what the question is asking. So far I have differentiated the function y and tried to work out the gradient when x has a value of pi/4 but I'm getting a weird decimal answer I think the question is asking me to work out the equation of the tangent to the curve and then sub in the values of when x=0 to work out the y intercept? Is this correct, and if not what am I doing wrong Thanks!
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    (Original post by Bugsy)
    Hey guys I'm just going through the uploaded question. I've managed part A, but on part B I'm unsure of what the question is asking. So far I have differentiated the function y and tried to work out the gradient when x has a value of pi/4 but I'm getting a weird decimal answer I think the question is asking me to work out the equation of the tangent to the curve and then sub in the values of when x=0 to work out the y intercept? Is this correct, and if not what am I doing wrong Thanks!
    Yes that is what it is asking, what have you got for the equation of the tangent?
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    (Original post by Bugsy)
    Hey guys I'm just going through the uploaded question. I've managed part A, but on part B I'm unsure of what the question is asking. So far I have differentiated the function y and tried to work out the gradient when x has a value of pi/4 but I'm getting a weird decimal answer I think the question is asking me to work out the equation of the tangent to the curve and then sub in the values of when x=0 to work out the y intercept? Is this correct, and if not what am I doing wrong Thanks!
    Find the equation of the tangent at  x = \dfrac{\pi}4

    Then sub in x=0 to get the answer.
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    (Original post by TheJ0ker)
    Yes that is what it is asking, what have you got for the equation of the tangent?
    I haven't managed to get an equation for the tangent yet, because I can't work out a value for m I tried to sub in pi/4 into my dy/dx but like I said I get a long decimal which I don't think is correct.. Once I get a M value I can use the y-y1=m(x-x1) to find the equation for the tangent..

    (Original post by raheem94)
    Find the equation of the tangent at  x = \dfrac{\pi}4

    Then sub in x=0 to get the answer.
    Yeah this is what I'm having problems with.. For my dy/dx I got 2x.Sec^2 x + 2tanx
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    (Original post by Bugsy)
    I haven't managed to get an equation for the tangent yet, because I can't work out a value for m I tried to sub in pi/4 into my dy/dx but like I said I get a long decimal which I don't think is correct.. Once I get a M value I can use the y-y1=m(x-x1) to find the equation for the tangent..



    Yeah this is what I'm having problems with.. For my dy/dx I got 2x.Sec^2 x + 2tanx
    Don't just plug the numbers into a calculator then. If you do it using your grey matter you will get a simple gradient involving an integer and pi.
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    (Original post by Bugsy)
    Yeah this is what I'm having problems with.. For my dy/dx I got 2x.Sec^2 x + 2tanx
    Sub in  x = \dfrac{\pi}4

     \displaystyle \frac{dy}{dx} = 2 \times \frac{\pi}4 \times sec^2\frac{\pi}{4} + 2 tan \frac{\pi}4 = \frac{\pi}2 \times 2 + 2 = 2 + \pi

    It is an exact value.
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    (Original post by TheJ0ker)
    Don't just plug the numbers into a calculator then. If you do it using your grey matter you will get a simple gradient involving an integer and pi.
    (Original post by raheem94)
    Sub in  x = \dfrac{\pi}4

     \displaystyle \frac{dy}{dx} = 2 \times \frac{\pi}4 \times sec^2\frac{\pi}{4} + 2 tan \frac{\pi}4 = \frac{\pi}2 \times 2 + 2 = 2 + \pi

    It is an exact value.
    Ah okay, great I get the answer now! Sorry for being so silly, I should have tried to do it without a calculator to get an exact value.. Thanks for your help

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