jmaths
IB Mathematics: Applications and Interpretation SL · 12.9: Differentiation · 12.9.5
Differentiating powers of x Problem set B One question at a time
Try each question on paper first. Open the working a line at a time, then say whether you had it.
Skills
-
Differentiate each function.
(a) \(g(x)=x^7\)
- \(g'(x)=7x^{6}\)
(b) \(z(t)=t^{-8}\)
- \(z'(t)=-8t^{-8-1}=-8t^{-9}\)
(c) \(p=q^{-7}\)
- \(\frac{dp}{dq}=-7q^{-8}\)
-
Differentiate each function.
(a) \(y=6x^5\)
- \(6\times5x^4=30x^4\)
(b) \(g(u)=-2u^7\)
- \(-2\times7u^6=-14u^6\)
(c) \(y=-8\)
- The derivative of a constant is 0
(d) \(y=3x-10\)
- The derivative of 3x is 3, and the derivative of −10 is 0
-
Rewrite as a power of the variable, then differentiate.
(a) \(y=\dfrac{7}{x^2}\)
- \(y=7x^{-2}\)
- \(\frac{dy}{dx}=7\times(-2)x^{-3}=-14x^{-3}\)
(b) \(y=-\dfrac{9}{t^4}\)
- \(y=-9t^{-4}\)
- \(\frac{dy}{dt}=-9\times(-4)t^{-5}=36t^{-5}\)
(c) \(f(x)=-\dfrac{5}{2x}\)
- \(f(x)=-\tfrac52x^{-1}\)
- \(f'(x)=-\tfrac52\times(-1)x^{-2}=\tfrac52x^{-2}\)
(d) \(y=\dfrac{1}{5x^5}\)
- \(y=\tfrac15x^{-5}\)
- \(\frac{dy}{dx}=\tfrac15\times(-5)x^{-6}=-x^{-6}\)
-
Differentiate each function.
(a) \(y=5x^3-8x^2+x-4\)
- Differentiate term by term: 15x², −16x, 1 and 0
(b) \(y=\tfrac34x^4-\tfrac25x^5\)
- \(\tfrac34\times4=3\)
- \(\tfrac25\times5=2\)
(c) \(f(x)=7-2x^2+\tfrac13x^3\)
- The constant 7 gives 0
- \(\tfrac13\times3=1\)
(d) \(y=x^3-2x+\dfrac6x\)
- \(y=x^3-2x+6x^{-1}\)
- \(\frac{dy}{dx}=3x^2-2-6x^{-2}\)
-
Expand the brackets, then differentiate.
(a) \(f(x)=x^3(5x-2)\)
- \(f(x)=5x^4-2x^3\)
- \(f'(x)=20x^3-6x^2\)
(b) \(g(x)=(x-6)(x+4)\)
- \(g(x)=x^2-2x-24\)
- \(g'(x)=2x-2\)
(c) \(h(x)=(2x-3)(x+5)\)
- \(h(x)=2x^2+7x-15\)
- \(h'(x)=4x+7\)
-
Expand the brackets, then differentiate.
(a) \(h(x)=\left(3+\dfrac1x\right)^2\)
- \(h(x)=9+6x^{-1}+x^{-2}\)
- \(h'(x)=-6x^{-2}-2x^{-3}\)
(b) \(g(x)=\left(2x-\dfrac5x\right)^2\)
- \(g(x)=4x^2-20+25x^{-2}\)
- The middle term is 2 × 2x × (−5/x) = −20
- \(g'(x)=8x-50x^{-3}\)
In context
-
A culture of bacteria has \(N=3t^4\) thousand cells after t hours. Find the rate of change of N, \(\frac{dN}{dt}\), when \(t=3\). Give the unit.
- \(\frac{dN}{dt}=12t^3\)
- \(t=3:\ 12\times 27=324\)
- The rate is 324 thousand cells per hour
-
The force of gravity on a probe is \(F=\dfrac{360}{r^3}\) newtons at a distance r thousand km from a planet. Find \(\frac{dF}{dr}\) when \(r=3\), to 3 significant figures.
- \(F=360r^{-3}\)
- \(\frac{dF}{dr}=-1080r^{-4}\)
- \(r=3:\ -\frac{1080}{81}=-13.3\)
- The force falls by 13.3 newtons per thousand km
-
The height of a model rocket is \(h=-\tfrac34t^2+18t+2\) metres, t seconds after launch.
(a) Find \(\frac{dh}{dt}\).
- \(-\tfrac34\times2=-\tfrac32\)
- \(\frac{dh}{dt}=-\tfrac32t+18\)
(b) Find the rate of change of height when \(t=4\).
- \(-\tfrac32\times4+18=12\)
(c) Find the value of t for which the rate of change is 0.
- \(-\tfrac32t+18=0\)
- \(t=12\)
Stretch
-
The function \(f(x)=px^3+qx\), where p and q are constants, has \(f'(1)=5\) and \(f'(2)=41\). Find p and q.
- \(f'(x)=3px^2+q\)
- \(f'(1)=3p+q=5\)
- \(f'(2)=12p+q=41\)
- Subtract: 9p = 36, so p = 4, then q = 5 − 12 = −7
-
The function \(g(x)=kx^n\), where k and n are constants, has derivative \(g'(x)=-20x^{-5}\). Find k and n.
- \(g'(x)=knx^{n-1}\)
- \(n-1=-5\Rightarrow n=-4\)
- \(kn=-20\Rightarrow -4k=-20\Rightarrow k=5\)
jmaths.xyz · AI SL · 12.9.5 Differentiating powers of x