# can someone pls help me solve this really long derivative?

### 2 Answers

- PhilipLv 61 month ago
y=(2/x^6)[400x^12+400x^5+(100/x^2)]^(1/2);

=2[400+400x^(-7)+100x^(-10)]^(1/2).;

Put u = 400+400x^(-7)+100x^(-10);

Then du/dx = -2800x^(-8)-1000x^(-11);

...................= -200x^(-11)[14x^3+5];

dy/dx = {d/dx}[2u^(1/2)] = [u^(-1/2)]{du/dx};

= -200x^(-11)[14x^3 +5]*[400+400x(-7)+100x^(-10)]^(-1/2)

- billrussell42Lv 71 month ago
y = (2/x⁶)√(400x¹² + 400x⁵ + (100/x²))

y = √(4/x¹²)√(400x¹² + 400x⁵ + (100/x²))

y = √(400 + 1600/x⁷ + 400/x¹⁴)

chain rule

y' = (dy/du)(du/dx)

dy/du = (1/2)(400 + 1600/x⁷ + 400/x¹⁴)^(–1/2)

du/dx = (–7)(1600)/x⁸ + (–14)(400)/x¹⁵ = –11200/x⁸ – 5600/x¹⁵

y' = [(1/2)(400 + 1600/x⁷ + 400/x¹⁴)^(–1/2)] [–11200/x⁸ – 5600/x¹⁵]

y' = –[11200/x⁸ + 5600/x¹⁵] / 2√(400 + 1600/x⁷ + 400/x¹⁴)y' = –[11200/x⁸ + 5600/x¹⁵] / 2(20)√(1 + 4/x⁷ + 1/x¹⁴)y' = –[11200/x⁸ + 5600/x¹⁵] / (40)√(1 + 4/x⁷ + 1/x¹⁴)y' = –[280/x⁸ + 140/x¹⁵] / √(1 + 4/x⁷ + 1/x¹⁴)y' = –(1/x⁸)[280 + 140/x⁷] / √(1 + 4/x⁷ + 1/x¹⁴)y' = –[280 + 140/x⁷] / x⁸√(1 + 4/x⁷ + 1/x¹⁴)y' = –[280 + 140/x⁷] / √(x¹⁶)√(1 + 4/x⁷ + 1/x¹⁴)y' = –[280 + 140/x⁷] / √(x¹⁶ + 4x⁹ + x²)y' = –[280 + 140/x⁷] / √x²√(x¹⁴ + 4x⁷ + 1)y' = –[280 + 140/x⁷] / x√(x¹⁴ + 4x⁷ + 1)