# Calculus

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what is the equation of the tangent line that passes through

X^(1/4) + y^(1/4) =4 at the point (16,16)

• Calculus -

(1/4)x^(-3/4) + (1/4)y^(-3/4) dy/dx = 0
dy/dx = -x^(-3/4)/y^(-3/4)
= y^(3/4)/x^(3/4)

sub in the point(16,16) to get the slope.
since you have a point on the line and the slope, it becomes an easy question.

btw, 16^(3/4) = [ 16^(1/4) ]^3 = 2^3 = 8
so slope = 8/8 = 1

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