Anti-differentiation: reversing the power rule Problem set A

Find f(x) in each case.

1. (a) \(f'(x)=x^{8}\)
2. (b) \(f'(x)=x^{-6}\)
3. (c) \(f'(x)=1\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=14x^{6}\)
2. (b) \(\frac{dy}{dx}=\frac{5}{2}x^{4}\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=-4x^{-5}\)
2. (b) \(\frac{dy}{dx}=\frac{3}{4}x^{-7}\)

Find y in each case.

1. (a) \(\frac{dy}{dx}=5x^{4} - 6x^{2} + 9\)
2. (b) \(\frac{dy}{dx}=\frac{3}{4}x^{2} + \frac{5}{3}x\)

Find \(\int\left(4x^{3} + 6x^{-3} - 5\right)dx\).

Find \(\int\left(\frac{5}{2}y^{-3} - \frac{4}{3}y^{-5}\right)dy\).

Rewrite as a power of x, then find f(x).

1. (a) \(f'(x)=-\frac{10}{x^{3}}\)
2. (b) \(f'(x)=\frac{12}{x^{5}}\)

Find f(x) in each case.

1. (a) \(f'(x)=9\sqrt{x}\)
2. (b) \(f'(x)=10x^{-1/2}\)

The velocity of a cyclist along a straight track is \(v=\frac{ds}{dt}=6t^{2}-10t+4\) m/s, where s is the displacement in metres from the start line and t the time in seconds.

1. (a) Find an expression for s in terms of t, including the constant c.
2. (b) The cyclist is at the start line when \(t=0\), so \(s=0\). Write down the value of c.

The strength S of a radio signal changes with the distance d km from the transmitter at a rate \(\frac{dS}{dd}=-\frac{12}{d^{3}}+2d\). Find an expression for S in terms of d.

Find f(x) when \(f'(x)=\frac{4}{x^{2}}+6\sqrt{x}-3x^{-1/2}\).

Sam writes: \(\int 6x^{-4}\,dx=-24x^{-5}+c\). Say what Sam did wrong and give the correct answer.