Question

DETERMINAR AS DERIVADAS POR REGRA DA CADEIA OU DERIVAÇÃO IMPLÍCITA.
OBS:

C) X²-Y²=0

Answer 1

Regra da cadeia:

$\boxed{\boxed{\dfrac{d}{dx}f(g(x))=\dfrac{dy}{du}\dfrac{du}{dx}=f'(g(x))\cdot g'(x)}}$
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A)

$y=(x+2)^{7}$

Seja f(x) = x⁷ e g(x) = x + 2, então, f(g(x)) = (g(x))⁷ = (x + 2)⁷

$f'(x)=7x^{6}\\g'(x)=1x^{1-1}+0=1$

Vamos derivar y utilizando as derivadas de f e g:

$y'=f'(g(x))\cdot g'(x)\\\\y'=f'(x+2)\cdot g'(x)\\\\y'=7(x+2)^{6}\cdot1\\\\\boxed{\boxed{y'=7(x+2)^{6}}}$

B)

$y=sen(x^{2})$

Seja f(x) = sen(x) e g(x) = x², então y = f(g(x)) = sen(g(x)) = sen(x²)

$f'(x)=cos(x)\\g'(x)=2x$

Derivando y:

$y'=f'(g(x))\cdot g'(x)\\\\y'=f'(x^{2})\cdot g'(x)\\\\\boxed{\boxed{y'=cos(x^{2})\cdot2x}}$

C)

$x^{2}-y^{2}=0$

Derivando implicitamente em relação a x:

$2x^{2-1}\dfrac{dx}{dx}-2y^{2-1}\dfrac{dy}{dx}=0\\\\\\2x-2y\dfrac{dy}{dx}=0\\\\\\x-y\dfrac{dy}{dx}=0\\\\\\y\dfrac{dy}{dx}=x\\\\\\\boxed{\boxed{\dfrac{dy}{dx}=\dfrac{x}{y}}}$