I need to know about the same time as the first time in the world of the most important thing is that the only one of the

Mathematics

1 answer:

Tresset [83]3 years ago

5 0

Cant think of something more logical than this

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Regard y as the independent variable and x as the dependent variable and use implicit differentiation to find dx/dy. x5y2 − x4y

il63 [147K]

Answer:

dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3).

Step-by-step explanation:

x^5y^2 − x^4y + 2xy^3 = 0

Applying the Product and Chain Rules:

y^2*5x^4*dx/dy + 2y*x^5 - (y*4x^3*dx/dy + x^4) + (y^3* 2*dx/dy + 3y^2*2x) =0

Separating the terms with derivatives:

y^2*5x^4*dx/dy - y*4x^3*dx/dy + y^3* 2*dx/dy = x^4 - 2y*x^5 - 3y^2*2x

dx/dy = (x^4 - 2x^5y - 6xy^2) / (5x^4y^2 - 4x^3y + 2y^3)

7 0

3 years ago

Read 2 more answers

If f(x)=log10x, what is the approximate value of y in the equation 10^y=5?

Simora [160]

$\bf log_{{ a}}\left( x^{{ b}} \right)\implies {{ b}}\cdot log_{{ a}}(x)\\\\
-----------------------------\\\\
10^y=5\implies log(10^y)=log(5)\implies ylog(10)=log(5)\\\\\\ y=\cfrac{log(5)}{log(10)}\\\\
-----------------------------\\\\
\textit{Logarithm Change of Base Rule}\\\\
log_{{ a}}{{ b}}\implies \cfrac{log_{{ c}}{{ b}}}{log_{{ c}}{{ a}}}\\\\
-----------------------------\\\\
y=\cfrac{log(5)}{log(10)}\iff y = log_{10}(5)\iff y=log(5)$