Question

Derive as funcoes implicita f (x)= x/x+1

Answer 1

Iremos aplicar a regra da derivada do quociente:

$\mathsf{\bold{\dfrac{f\left(x\right)}{g\left(x\right)}=\dfrac{f'\left(x\right)\cdot g\left(x\right)-f\left(x\right)\cdot g'\left(x\right)}{\left[g\left(x\right)\right]^2}}}$

A função em questão é:

$\mathsf{f\left(x\right)=\dfrac{x}{x+1}}$

Aplicando a regra teremos:

$\mathsf{f'\left(x\right)=\dfrac{\left[1\cdot \:\left(x+1\right)\right]-\left(x\cdot 1\right)}{\left(x+1\right)^2}}\\\\\\\mathsf{f'\left(x\right)=\dfrac{x+1-x}{\left(x+1\right)^2}}\\\\\\\mathsf{\boxed{\bold{f'\left(x\right)=\dfrac{1}{\left(x+1\right)^2}=\dfrac{1}{x^2+2x+1}}}}}$

Answer 2

$\mathsf{f(x)=\dfrac{x}{x+1}}\\\mathsf{f'(x)=\dfrac{1\cdot(x+1)-x\cdot1}{(x+1)^2}}\\\mathsf{f'(x)=\dfrac{x+1-x}{(x+1)^2}}\\\huge\boxed{\boxed{\boxed{\boxed{\mathsf{f'(x)=\dfrac{1}{(x+1)^2}}}}}}$

$\dotfill$

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