Introducing a Useful Dodge

Excerpts

• Introducing a Useful Dodge

  Thus, the equationdodge y = (x^2+a^2)^32 is awkward to a beginner.

• Introducing a Useful Dodge

  Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.

• Introducing a Useful Dodge

  By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will find this dodge of increasing usefulness.

• Introducing a Useful Dodge

  The process can be extended to three or more differential coefficients, so that $\dfrac{dy}{dx} = \dfrac{dy}{dz} × \dfrac{dz}{dv} × \dfrac{dv}{dx}$.

• Introducing a Useful Dodge

  (We may also write $y = (1-x)^{\efrac{1}{2}} (1+x)^{-\efrac{1}{2}}$ and differentiate as a product.)

• Introducing a Useful Dodge

  (1) Differentiate $y = \sqrt{a+x}$. Let $a+x = u$.

Equations

• Introducing a Useful Dodge

  \frac{dy}{du} = \frac{3}{2} u^{\efrac{1}{2}}

  Differentiating y = u^(3/2) with respect to u gives three halves times u to the power one half.

• Introducing a Useful Dodge

  \frac{du}{dx} = 2x

  The derivative of u = x^2 + a^2 with respect to x is 2x.

• Introducing a Useful Dodge

  \frac{dy}{dx} = \frac{dy}{du}×\frac{du}{dx}

  The derivative of y with respect to x is the product of the derivative of y with respect to u and the derivative of u with respect to x.

• Introducing a Useful Dodge

  &= 3x(x^2 + a^2)^{\efrac{1}{2}}

  The derivative of (x^2 + a^2)^(3/2) with respect to x is 3x times (x^2 + a^2) to the power one half.

• Introducing a Useful Dodge

  \frac{dy}{dx} &= \frac{dy}{du} × \frac{du}{dx} = \frac{1}{2\sqrt{a+x}}

  The derivative of y = sqrt(a+x) with respect to x is 1 over twice the square root of a+x.

• Introducing a Useful Dodge

  \frac{dy}{dx} &= \frac{dy}{du}×\frac{du}{dx} = - \frac{x}{\sqrt{(a+x^2)^3}}

  The derivative of y = 1/sqrt(a+x^2) with respect to x is minus x over the square root of (a+x^2) cubed.

• Introducing a Useful Dodge

  \frac{dy}{dx} &= \frac{dy}{du} × \frac{du}{dx} = -\frac{3x^2}{2\sqrt{(x^3 - a^2)^3}}

  The derivative of y = 1/sqrt(x^3 - a^2) with respect to x is minus 3x^2 over twice the square root of (x^3 - a^2) cubed.

• Introducing a Useful Dodge

  \frac{dy}{dx} &= - \frac{1}{(1+x)\sqrt{1-x^2}}

  The derivative of y = sqrt((1-x)/(1+x)) with respect to x is minus 1 over (1+x) times the square root of (1-x^2).

• Introducing a Useful Dodge

  = \frac{\sqrt{x}(3+x^2)}{2\sqrt{(1+x^2)^3}}

  The derivative of y = sqrt(x^3/(1+x^2)) with respect to x is sqrt(x) times (3+x^2), over twice the square root of (1+x^2) cubed.

• Introducing a Useful Dodge

  &= 3\left(x+\sqrt{x^2+x+a}\right)^2 \left(1 +\frac{2x+1}{2\sqrt{x^2+x+a}}\right)

  The derivative of y = (x + sqrt(x^2+x+a))^3 with respect to x is three times the square of (x + sqrt(x^2+x+a)), times (1 + (2x+1) over twice sqrt(x^2+x+a)).

• Introducing a Useful Dodge

  = \frac{x(3a-4x)}{2b\sqrt{(a-x)x}}

  The first derivative of y = (x/b) sqrt((a-x)x) with respect to x is x(3a-4x) over 2b times the square root of (a-x)x.

• Introducing a Useful Dodge

  &= \frac{3a^2-12ax+8x^2}{4b(a-x)\sqrt{(a-x)x}}

  The second derivative of y = (x/b) sqrt((a-x)x) with respect to x is (3a^2 - 12ax + 8x^2) over 4b(a-x) times the square root of (a-x)x.

• Introducing a Useful Dodge

  \frac{d(y^n)}{d(y^5)} = \frac{ny^{n-1}}{5y^{5-1}} = \frac{n}{5} y^{n-5}

  The derivative of y^n with respect to y^5 equals n/5 times y to the power n minus 5.

• Introducing a Useful Dodge

  \dfrac{dy}{dx} = \dfrac{dy}{dz} × \dfrac{dz}{dv} × \dfrac{dv}{dx}

  The chain rule extends to any number of intermediate quantities, so dy/dx is the product of successive derivatives along the chain of variables.

• Introducing a Useful Dodge

  \frac{dv}{dx} = \frac{7x(5x-6)}{3\sqrt[3]{(x-1)^4}}

  The derivative of v = 7x^2 / cube root of (x-1) with respect to x is 7x(5x-6) over 3 times the cube root of (x-1)^4.

• Introducing a Useful Dodge

  \frac{dx}{dt} = 3t^2 + \tfrac{1}{2}

  The derivative of x = t^3 + t/2 with respect to t is 3t^2 + 1/2.

• Introducing a Useful Dodge

  \frac{dt}{d\theta} = -\frac{1}{10\sqrt{\theta^3}}

  The derivative of t = 1/(5 sqrt(theta)) with respect to theta is minus 1 over 10 times the square root of theta cubed.

• Introducing a Useful Dodge

  \frac{dv}{d\theta} = -\frac{7x(5x-6)(3t^2+\frac{1}{2})} {30\sqrt[3]{(x-1)^4} \sqrt{\theta^3}}

  The derivative of v with respect to theta, by the chain of intermediate variables, is minus 7x(5x-6)(3t^2 + 1/2) over 30 times the cube root of (x-1)^4 times the square root of theta cubed.

• Introducing a Useful Dodge

  = -\frac{28}{3x^5\sqrt{9x^8+7}}

  For y = sqrt(1+v), v = 7/z^2, z = 3x^4, the book's chain of derivatives gives this value, which it labels dy/dx although the question asks for dv/dx.

• Introducing a Useful Dodge

  \frac{d\phi}{d\omega} = \frac{1}{\sqrt{2}\omega^2}

  The derivative of phi = sqrt(3) - 1/(omega sqrt(2)) with respect to omega is 1 over sqrt(2) times omega squared.

• Introducing a Useful Dodge

  \frac{d\omega}{d\theta} = -\frac{1}{(1+\theta)\sqrt{1-\theta^2}}

  The derivative of omega with respect to theta is minus 1 over (1+theta) times the square root of (1-theta^2).

Problems