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3 years ago

HELP!!!! THIS IS DUE TOMORROW:(((

Mathematics

2 answers:

Verdich [7]3 years ago

6 0

This is actually really simple algebra. All you need to know is that Y= Mx + B
and M BEING YOUR SLOPE good luck

notsponge [240]3 years ago

3 0

The answer is A.

4x +y=-9
y= -4x - 9

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lidiya [134]

Answer:

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}$

Step-by-step explanation:

The Derivative of a Function

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

$\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

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$u=(x+\sqrt{1+x^2})$

Then

$\displaystyle u'=1+\frac{2x}{2\sqrt{1+x^2}}$

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Taking the derivative of y

$y'=3[tan^{-1}(x+\sqrt{1+x^2})]'$

Using the change of variables

$\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}$

$\displaystyle y'=3\frac{u'}{1+u^2}=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+(x+\sqrt{1+x^2})^2}$

Operating

$\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{1+x^2+2x\sqrt{1+x^2}+1+x^2}$

$\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}$

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Step-by-step explanation:

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