Question

calcular a derivada 3° ordem da função f(x)= (2x-3)⁸ no ponto X0=1​

Answer 1

$\mathsf{f(x)=(2x-3)^8}\\\mathsf{f'(x)=8(2x-3)^7.2=16(2x-3)^7}$

$\mathsf{f''(x)=112(2x-3)^6.2=224(2x-3)^6}$

$\mathsf{f'''(x)=1334.(2x-3)^5.2}\\\mathsf{f'''(x)=2688(2x-3)^5}$

$\mathsf{f'''(1)=2688.(2.1-3)^5=2688.(-1)=-2688}$