[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?

[yusufu]

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.

[Gaz031]

Originally Posted by 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?

The question relies on your know the derivative of [Mimetex cannot convert this formula] - you can derive the result by writing [Mimetex cannot convert this formula] then using the derivative of [Mimetex cannot convert this formula] which you should know.
Find the gradient and then apply the standard equation of a line [Mimetex cannot convert this formula]

[Dekota]

Originally Posted by 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