The Student Room Group
Reply 1
differentiate it then find the gradient where x=2.
then find the equation of the straight line through the point (2,4.25) with the specified gradient.
Reply 2
bigbadb1319
Ex. 4C
Q2. Find the equation of the tangent to the curve y=2^x + 2^(-x) at the point (2, 4.25)

I know you have to differentiate this, but what do you do after that? :confused:

The question relies on your know the derivative of axa^{x} - you can derive the result by writing ax=(elna)x=exlnaa^{x}=(e^{\ln a})^{x}=e^{x\ln a} then using the derivative of ekxe^{kx} which you should know.
Find the gradient and then apply the standard equation of a line yy1=m(xx1)y-y_{1}=m(x-x_{1})
Reply 3
bigbadb1319
Find the equation of the tangent to the curve y=2^x + 2^(-x) at the point (2, 4.25)


y = 2^x + 2^(-x)
-> dy/dx = d[2^x]/dx + d[2^(-x)]/dx
............ = d[2^x]/dx + 2^(-x)ln2.d[-x]/dx
............ = 2^x.ln2 - 2^(-x).ln2
............ = 2^(-x).(4^x - 1)ln2

sub (x = 2) into dy/dx ... 2^(-2).(4² - 1)ln2 ≈ 2.6

sub (y = 4.25, x = 2, dy/dx = 2.6) into straight line equation:

-> y - 4.25 = 2.6(x - 2)
-> y = 2.6x - 0.948

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