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can someone help me with some calculus differentiation? tangents to a curve?
FInd the equation of the tangent to the curve y^4-x^4=15 at the point (1,2).
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- WoodsmanLv 79 years agoFavourite answer
For y = (15 + x^4)^(1/4),
dy/dx = x^3(15 + x^4)^(-3/4)
See cite. Click "show steps" if you like.
At x=1, dy/dx = 16^(-3/4) = 1/8
y = mx + b
2 = (1/8)1 + b
b = 15/8, so the equation of the tangent at that point is
y = x/8 + 15/8
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