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zaharov [31]
3 years ago
5

I need help solving this problem.

E%7B2%7D%20%7D%2C%20" id="TexFormula1" title="4x cos ^{-1} (2x+4)- \sqrt{3-3 x^{2} }, " alt="4x cos ^{-1} (2x+4)- \sqrt{3-3 x^{2} }, " align="absmiddle" class="latex-formula"> find f'(x). My original which is wrong, is included.

Mathematics
2 answers:
Nookie1986 [14]3 years ago
8 0
I believe it's as follows:

f'(x) = 4(acos(2x+4)- \frac{2x}{ \sqrt{-4x^2-16x-15}})+ \frac{3x}{ \sqrt{3-3x^2}}
Gnesinka [82]3 years ago
3 0
Given:
f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}

Using
\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}
we derive
\frac{d}{dx}4xcos^{-1}(2x+4)
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}

Similarly, using
\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}
we derive
\frac{d}{dx}(-\sqrt{3-3x^2})
=\frac{3x}{\sqrt{3-3x^2}}

Therefore, the derivative is
f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}
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