Answer: (a) \(\displaystyle \frac{10(4x-2)^4}{\sqrt{3+(4x-2)^5}}\); (b) \(0.00016\); (c)(i) \(x^2(1+3\ln x)\); (c)(ii) \(\displaystyle \frac{x^3}{18}(1+3\ln x)+C\).

Differentiate first. Stationary points occur where the derivative is zero, while tangent and normal problems use the gradient at the given point.

(a) Write

\(f(x)=\bigl(3+(4x-2)^5\bigr)^{1/2}.\)

Using the chain rule,

\(f'(x)=\frac12\bigl(3+(4x-2)^5\bigr)^{-1/2}\cdot5(4x-2)^4\cdot4.\)

Therefore

\(f'(x)=\frac{10(4x-2)^4}{\sqrt{3+(4x-2)^5}}.\)

(b) Differentiate

\(y=\frac{5x}{3x+2}.\)

Using the quotient rule,

\(\frac{dy}{dx}=\frac{(3x+2)(5)-5x(3)}{(3x+2)^2}.\)

So

\(\frac{dy}{dx}=\frac{10}{(3x+2)^2}.\)

When \(y=10\),

\(10=\frac{5x}{3x+2}.\)

Hence

\(30x+20=5x,\)

so

\(x=-\frac45.\)

At this value of \(x\),

\(3x+2=-\frac25.\)

Therefore

\(\frac{dy}{dx}=\frac{10}{( -2/5)^2} =\frac{10}{4/25} =62.5.\)

For small changes,

\(\Delta y\approx\frac{dy}{dx}\Delta x.\)

Thus

\(0.01\approx62.5\Delta x.\)

So

\(\Delta x\approx\frac{0.01}{62.5}=0.00016.\)

(c)(i) Using the product rule,

\(\frac{d}{dx}(x^3\ln x)=3x^2\ln x+x^3\cdot\frac1x.\)

So

\(\frac{d}{dx}(x^3\ln x)=3x^2\ln x+x^2=x^2(1+3\ln x).\)

(c)(ii) Rewrite the integrand as

\(\frac{x^2}{6}(2+3\ln x) =\frac16x^2(1+3\ln x)+\frac16x^2.\)

Using part (c)(i),

\(\int\frac16x^2(1+3\ln x)\,dx=\frac16x^3\ln x.\)

Also,

\(\int\frac16x^2\,dx=\frac{x^3}{18}.\)

Therefore

\(\int\frac{x^2}{6}(2+3\ln x)\,dx =\frac{x^3}{6}\ln x+\frac{x^3}{18}+C.\)

Equivalently,

\(\frac{x^3}{18}(1+3\ln x)+C.\)