(a) Find the derivative of \(4x-\frac{7}{x^2}\)with respect to \(x\), from first principle. (b) Given that tan \(P =\frac{3}{x - 1}\) and tan \(Q\) =\frac{2...

Answer Details

(a) \(y=4x-\frac{7}{x^2}\)

\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}\)

\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}-(4x-\frac{7}{x^2})\)

\(dy=4dx-\frac{7}{(x + dx)^2}+\frac{7}{x^2}\)

\(dy=4dx-(\frac{7}{(x + dx)^2}-\frac{7}{x^2})\)

\(dy=4dx-\frac{7x^2 - 7(x + dx)^2}{x^2(x + dx)^2)}\)

\(dy=4dx-\frac{(7(x^2 - (x + dx)^2)}{x^2(x + dx)^2)}\)

\(a^a-b^2=(a+b)(a-b)\)

\(dy=4dx-(\frac{7((x+x+dx)(x-x-dx))}{x^2(x+dx)^2})\)

\(dy=4dx-(\frac{7((2x+dx)(-dx))}{x^2(x+dx)^2})\)

\(dy=dx(4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2}))\)

\(dy=4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2})\)

As \(dx \to 0 \frac{dy}{dx}=4-(\frac{7((2x)(-1))}{x^2(x)^2})\)

\(\frac{dy}{dx}=4-(\frac{-14x}{x^4}\)

\(\therefore \frac{dy}{dx}=4+\frac{14}{x^3}\)

(b) tan \(P= \frac{3}{x - 1},tan Q={2}{x + 1}\)

\(tan (P-Q)=\frac{tan P - tan Q}{1 + tan P tan Q}\)

tan P-tan Q=\(\frac{3}{x - 1}-\frac{2}{x + 1}=\frac{3(x + 1) - 2(x - 1)}{(x + 1)(x - 1)}\)

\(=\frac{3x + 3 - 2x + 2}{(x + 1)(x - 1)}=\frac{x + 5}{(x + 1)(x - 1)}\)

1+tan P tan Q =\(1+(\frac{3}{x - 1})(\frac{2}{x + 1})=1+\frac{6}{(x + 1)(x - 1)}\)

=\(\frac{(x + 1)(x - 1) + 6}{(x + 1)(x - 1)}\)

\(\frac{tan P - tan Q}{1 + tan Ptan Q}=\frac{x + 5}{(x + 1)(x - 1)}÷\frac{(x + 1)(x - 1)+6}{(x + 1)(x - 1)}\)

\(=\frac{x + 5}{(x + 1)(x - 1)}\times\frac{(x + 1)(x - 1)}{(x + 1)(x - 1) + 6}\)

\(=\frac{x + 5}{(x + 1)(x - 1) + 6}=\frac{x + 5}{x^2 - 1 + 6}\)

∴tan(P-Q)\(=\frac{x + 5}{x^2 + 5}\)