Question

f(x)=\frac{2-x}{3x+1} find f'(x)

Derivatives
ANSWERED
asked 2021-05-13
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{2}-{x}}}{{{3}{x}+{1}}}}\) find f'(x)

Answers (1)

2021-05-14
Step 1
Consider the function:
\(\displaystyle{f{{\left({x}\right)}}}={\frac{{{2}-{x}}}{{{3}{x}+{1}}}}\)
Step 2
Differentiate the given function with respect to “x”
Apply quotient rules of derivatives,
\(\displaystyle{f}'{\left({x}\right)}={\frac{{{\left({3}{x}+{1}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({2}-{x}\right)}-{\left({2}-{x}\right)}{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{\left({3}{x}+{1}\right)}}}{{{\left({3}{x}+{1}\right)}^{{{2}}}}}}\)
\(\displaystyle={\frac{{{\left({3}{x}+{1}\right)}{\left({0}-{1}\right)}-{\left({2}-{x}\right)}{\left({3}+{0}\right)}}}{{{\left({3}{x}+{1}\right)}^{{{2}}}}}}\)
\(\displaystyle={\frac{{-{3}{x}-{1}-{6}+{3}{x}}}{{{\left({3}{x}+{1}\right)}^{{{2}}}}}}\)
\(\displaystyle{f}'{\left({x}\right)}=-{\frac{{{7}}}{{{\left({3}{x}+{1}\right)}^{{{2}}}}}}\)
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