Further Derivatives

Have a go first. Then open the working one line at a time.

1. Differentiate

   \(y'=\cos x+4e^x\)

   Differentiate term by term. A constant multiplies through.

2. Gradient at x = 0

   \(m=\cos 0+4e^0=1+4=5\)

   Put x = 0 in the derivative, not in the curve.

3. Equation

   \(y-4=5(x-0)\), so \(y=5x+4\)

   Use the point (0, 4) and \(y-y_1=m(x-x_1)\).

4. Check on a graph

   Desmos: f(x) = sin x + 4e^x, a = 0 gives m = f′(0) = 5

   The tangent drawn on the curve has the same gradient and passes through (0, 4).

Have a go first. Then open the working one line at a time.

1. Gradient needed

   The line has \(m=-1\). The tangent has \(m=1\).

   Negative reciprocal: \(-1\times 1=-1\).

2. Solve for x

   \(y'=2e^x-3=1\), so \(e^x=2\) and \(x=\ln 2=0.693\)

   Set the derivative equal to the gradient you need.

3. The point

   \(y=2e^{\ln 2}-3\ln 2=4-3\ln 2=1.92\)

   Put x back into the curve, not the derivative.

4. Equation

   \(y-(4-3\ln 2)=1(x-\ln 2)\), so \(y=x+4-4\ln 2\) (\(y=x+1.23\))

   Exact form first, then 3 significant figures.

Have a go first. Then open the working one line at a time.

1. Tangent gradient

   \(y'=4-e^x=4-\frac{17}{4}=-\frac14\)

   Put \(e^{\ln(17/4)}=\frac{17}{4}\) into the derivative.

2. Normal gradient and point

   \(m=4\) and \(y=4\ln\frac{17}{4}-\frac{17}{4}\)

   Negative reciprocal of \(-\frac14\) is 4.

3. Equation

   \(y=4x-\frac{17}{4}\)

   \(y-y_1=4(x-\ln\frac{17}{4})\): the \(4\ln\frac{17}{4}\) terms cancel.

4. Show: set equal

   \(4x-e^x=4x-\frac{17}{4}\), so \(e^x=\frac{17}{4}\)

   The 4x terms cancel.

5. Conclude

   \(e^x\) takes each value once, so \(x=\ln\frac{17}{4}\) is the only solution.

   One solution: the point you started from. The normal meets the curve nowhere else.

Have a go first. Then open the working one line at a time.

1. Split into powers

   \(h=6t^{1/2}+24t^{-1/2}\)

   Divide each term of the top by \(\sqrt{t}\).

2. Differentiate

   \(h'=3t^{-1/2}-12t^{-3/2}\)

   \(6\times\frac12=3\) and \(24\times(-\frac12)=-12\).

3. At t = 1

   \(h'(1)=3-12=-9\) m/s

   Give the units.

4. Meaning

   The height is decreasing at 9 metres per second.

   A negative rate means the bird is going down.

Have a go first. Then open the working one line at a time.

1. Tangent at x = a

   \(y=\sqrt a+\frac{1}{2\sqrt a}(x-a)\)

   The derivative of \(\sqrt x\) is \(\frac{1}{2\sqrt x}\). The point of tangency is \((a,\sqrt a)\).

2. Pass through (−2, 0)

   \(0=\sqrt a+\frac{-2-a}{2\sqrt a}\)

   Put x = −2 and y = 0 into the tangent.

3. Solve for a

   \(2a=a+2\), so \(a=2\)

   Multiply by \(2\sqrt a\).

4. The point

   \((2,\sqrt2)=(2,1.41)\)

   Check: gradient \(\frac{1}{2\sqrt2}\), and from (−2, 0) to (2, 1.41) the gradient is \(\frac{\sqrt2}{4}\). They match.

Have a go first. Then open the working one line at a time.

1. Match the gradient

   \(\frac{1}{2\sqrt x}=\frac14\)

   The tangent's gradient is \(\frac14\).

2. Solve

   \(\sqrt x=2\), so \(x=4\)

   Square both sides.

3. The point

   \(y=\sqrt4=2\), so the point is \((4,2)\)

   Check: the line gives \(\frac44+1=2\). It passes through the point.

Have a go first. Then open the working one line at a time.

1. Find P and Q

   \(6\cos x=0\) gives \(P\left(-\frac\pi2,0\right)\), \(Q\left(\frac\pi2,0\right)\)

   Radians. Base \(PQ=\pi\).

2. Gradients

   \(y'=-6\sin x\): at P, \(m=6\); at Q, \(m=-6\)

   \(\sin(-\frac\pi2)=-1\).

3. Tangents

   \(y=6\left(x+\frac\pi2\right)\) and \(y=-6\left(x-\frac\pi2\right)\)

   Each uses its own point.

4. Meet at R

   They meet where \(x=0\): \(R(0,3\pi)\).

   By symmetry R is on the y-axis. Height \(=3\pi\).

5. Area

   \(\frac12\times\pi\times3\pi=\frac{3\pi^2}{2}=14.8\)

   \(\frac12\times\text{base}\times\text{height}\).