Find the derivative of each function Choose 1 to work fx 2x

Find the derivative of each function ((Choose 1 to work!) f(x) = [(2x + 5) (x^3 - 2x)]^7 f(x) = (2x^2 + 1/3x + 1)^7

Solution

a) given f(x)=[(2x+5)(x²-2x)]⁷

differentiate with respect to x :

d/dx (f(x))ⁿ =n(f(x))^n-1*f \'(x) , product rule:(uv)\'=u\'v +uv\'

f \'(x)=7[(2x+5)(x²-2x)]^7-1[(2+0)(x²-2x)+(2x+5)(2x-2)]

f \'(x)=7[(2x+5)(x²-2x)]⁶[2x²-4x+ (2x+5)(2x-2)]

f \'(x)=7[(2x+5)(x²-2x)]⁶[2x²-4x+ 4x²+6x-10]

f \'(x)=7[(2x+5)(x²-2x)]⁶[6x²+2x-10]

f \'(x)=14[3x²+x-5][(2x+5)(x²-2x)]⁶

b)f(x)=[(2x²+1)_/(3x+1)]⁷

differentiate with respect to x :

d/dx (f(x))ⁿ =n(f(x))^n-1*f \'(x) , quotient rule:(u/v)\'=(u\'v -uv\')/v²

f \'(x)=7[(2x²+1)_/(3x+1)]^7-1[(4x+0)(3x+1) -(2x²+1)(3+0)]_/(3x+1)²

f \'(x)=7[(2x²+1)_/(3x+1)]⁶[12x²+4x -6x²-3]_/(3x+1)²

f \'(x)=7[(2x²+1)_/(3x+1)]⁶[6x²+4x -3]_/(3x+1)²

f \'(x)=7[6x²+4x -3][(2x²+1)⁶_/(3x+1)⁸]