Question

Determine a derivada dy/dx de:
7y^4 - 2y - 4x^3 = lnx + e^x

Answer 1

$7y^4 - 2y - 4x^3 = lnx + e^x\\\\7y^4 -2y = lnx +e^x +4x^3$

multiplicando os dois lados por d/dx e aplicando a distributiva

$\frac{d}{dx}[7y^4-2y] = \frac{d}{dx}[lnx+e^x+4x^3] \\\\ \frac{d}{dx}(7y^4) +\frac{d}{dx}(-2y) = \frac{d}{dx}(lnx) + \frac{d}{dx}(e^x)+ \frac{d}{dx}(4x^3)$

quando vc derivar o y em cima fica dy/dx...quando vc derivar o x fica dx/dx

$\frac{dy}{dx}(7*4y^{4-1}) +\frac{dy}{dx}(-2*1) = \frac{dx}{dx}( \frac{1}{x} ) + \frac{dx}{dx}(e^x)+ \frac{dx}{dx}(4*3x^{3-1})\\\\ \boxed{\boxed{\frac{dy}{dx} (28y^3-2)= \frac{dx}{dx}( \frac{1}{x}+e^x+12x^2 ) }}$

dx/dx = 1
isolando dy/dx passando 28y³-2 pro outro lado dividindo

${\frac{dy}{dx} (28y^3-2)=1( \frac{1}{x}+e^x+12x^2 ) }\\\\\\ \boxed{\boxed{\frac{dy}{dx}= ( \frac{1}{x}+e^x+12x^2 ) *( \frac{1}{28y^3-2} ) }}$