Asked by help1
Find the equation of the tangent line to the curve y = sqrt (x+5) at the point (4,3)
Answers
Answered by
Reiny
dy/dx = (1/2(x+5)^(-1/2)
or
= 1/(2√(x+5) )
at (4,3)
dy/dx = 1/(2√9) = 1/6
y-3 = (1/6)(x-4)
6y-18 = x-4
x - 6y = -14
or y = (1/6)x + 7/3
or
= 1/(2√(x+5) )
at (4,3)
dy/dx = 1/(2√9) = 1/6
y-3 = (1/6)(x-4)
6y-18 = x-4
x - 6y = -14
or y = (1/6)x + 7/3
Answered by
uh
Thanks! Sorry to bother you but do you know how I can get the same answer by solving using the definition of the derivative instead?
Answered by
Reiny
I used the derivative.
Do you want to find the derivative by "first principles" ?
Do you want to find the derivative by "first principles" ?
Answered by
uh
Oh no that's okay
The steps you put include the chain rule version of the power rule right?
The steps you put include the chain rule version of the power rule right?
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