Title: InkMaths / Differentiation: standard results
On every note of this deck, unless a note says otherwise:
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Front: STANDARD RESULT — LEARN IT
Natural log:
ddx(ln x) = ?\(\frac{d}{dx}(\text{ln } x) = ?\)(d/dx)(ln x) = ?
Back: ddx(ln x) = 1x\(\frac{d}{dx}(\text{ln } x) = \frac{1}{x}\)(d/dx)(ln x) = 1/x

Front: RECALL
State how to connect two rates of change.
Back: String the derivatives together with the chain rule so the middle variable cancels. To get dV/dt from a known dr/dt, write dV/dt = dV/dr × dr/dt. Write the chain you want first and then hunt for the pieces, rather than differentiating everything in sight.

Front: STANDARD RESULT — LEARN IT
Sine (x in radians):
ddx(sin x) = ?\(\frac{d}{dx}(\sin x) = ?\)(d/dx)(sin x) = ?
Back: ddx(sin x) = cos x\(\frac{d}{dx}(\sin x) = \cos x\)(d/dx)(sin x) = cos x

Front: STANDARD RESULT — LEARN IT
Power rule:
ddx(x^n) = ?\(\frac{d}{dx}(x^{n}) = ?\)(d/dx)(x^n) = ?
Back: ddx(x^n) = nx^n−1\(\frac{d}{dx}(x^{n}) = nx^{n−1}\)(d/dx)(x^n) = nx^(n − 1)

Front: STANDARD RESULT — LEARN IT
Exponential:
ddx(e^kx) = ?\(\frac{d}{dx}(e^{kx}) = ?\)(d/dx)(e^kx) = ?
Back: ddx(e^kx) = ke^kx\(\frac{d}{dx}(e^{kx}) = ke^{kx}\)(d/dx)(e^kx) = ke^kx

Front: STANDARD RESULT — LEARN IT
Implicit and parametric:
ddx(y^2) = ?, parametric dydx = ?\(\frac{d}{dx}(y^{2}) = ?, \text{parametric } \frac{dy}{dx} = ?\)(d/dx)(y^2) = ?, parametric (dy/dx) = ?
Back: 2ydydx, dydt ÷ dxdt\(2y\frac{dy}{dx}, \frac{dy}{dt} \div \frac{dx}{dt}\)2y(dy/dx), (dy/dt) ÷ dx/dt

Front: DEFINE
Define: differentiation from first principles.
Back: The derivative of f(x) is the limit as h tends to zero of (f(x + h) − f(x))/h, provided the limit exists.

Front: STANDARD RESULT — LEARN IT
Cosine (x in radians):
ddx(cos x) = ?\(\frac{d}{dx}(\cos x) = ?\)(d/dx)(cos x) = ?
Back: ddx(cos x) = −sin x\(\frac{d}{dx}(\cos x) = −\sin x\)(d/dx)(cos x) = − sin x

Front: CHECK YOURSELF
Differentiate y = 3 ln x − cos 2x.
Back: dy/dx = 3/x + 2 sin 2x.
The 2 appears from the chain rule on cos 2x, and the sign flips because the derivative of cos is minus sin.

Front: STANDARD RESULT — LEARN IT
Chain rule:
y in terms of u, u in terms of x: dydx = ?\(y \text{ in terms of } u, \, u \text{ in terms of } x: \frac{dy}{dx} = ?\)y in terms of u, u in terms of x: (dy/dx) = ?
Back: dydx = dydu × dudx\(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\)dy/dx = dy/du × du/dx

Front: RECALL
State how to classify a stationary point.
Back: Solve f'(x) = 0, then look at f''(x) there. Negative means a maximum, positive means a minimum. Zero settles nothing, so go back to the sign of f'(x) either side. Markers give no credit for a verdict with no second derivative and no sign table behind it.

Front: SPOT THE ERROR
Say what is wrong with this claim:
The derivative of cos x is sin x.
Back: It is −sin x. Sine's derivative is cos x with no sign change; cosine's picks up the minus. Sketch both curves: cos x is falling just after zero, so its gradient there must be negative.

Front: STANDARD RESULT — LEARN IT
General exponential:
ddx(a^kx) = ?\(\frac{d}{dx}(a^{kx}) = ?\)(d/dx)(a^kx) = ?
Back: ddx(a^kx) = ka^kx ln a\(\frac{d}{dx}(a^{kx}) = ka^{kx} \text{ln } a\)(d/dx)(a^kx) = ka^kx ln a

Front: STANDARD RESULT — LEARN IT
Product rule:
y = uv: dydx = ?\(y = uv: \frac{dy}{dx} = ?\)y = uv: (dy/dx) = ?
Back: dydx = udvdx + vdudx\(\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}\)dy/dx = u(dv/dx) + v(du/dx)