Math Is Fun Forum

ashleyrc

Member

Registered: 2008-10-06

Posts: 3

derivative problem (down to simplifying?)

find equation of the tangent line to the graph of the function f(x)=square root of x^2+16, when x = 3/

what i did: think i'm supposed to just take the derivative, which would equal f(x) = x + 4 ; f'(x) = 1? does that mean limit does not exist?

ashleyrc

Member

Registered: 2008-10-06

Posts: 3

Re: derivative problem (down to simplifying?)

sorry, not when x = 3/. i meant to say when x=3

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: derivative problem (down to simplifying?)

It's not that the tangent line doesn't exist, it's just that it's exactly the same as the function, because the function is just a line as well.

Except it's not. When taking the square root of things, you're not allowed to just take the square root of each individual term.
Using the chain rule to find the derivative gets that

.

The gradient of the tangent would then be 3/5, and you can find the y-intercept by using that gradient with the point (3,5).

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Why did the vector cross the road?
It wanted to be normal.