Tag: Chain Rule

Compute the derivative of the given function

by RoRi

September 21, 2015

Compute the derivative of the function

$f(x) = x \sqrt{1+x^2}.$

Using the chain rule,

$\begin{align*} f'(x) &= \sqrt{1+x^2} + x (2x)\left(\frac{1}{2}\right)(1+x^2)^{-\frac{1}{2}} \\[8pt] &= \frac{1+2x^2}{\sqrt{1+x^2}}. \end{align*}$

Compute the derivative of the given function

by RoRi

September 21, 2015

Compute the derivative of the function

$f(x) = \sec^2 x + \csc^2 x.$

First, we simplify the given expression,

$\begin{align*} f(x) = \sec^2 x + \csc^2 x &= \frac{1}{\cos^2 x} + \frac{1}{\sin^2 x} \\ &= \frac{\sin^2 x + \cos^2 x}{\cos^2 x \cdot \sin^2 x} \\ &= \frac{1}{\frac{1}{4} \sin^2 (2x)} \\ &= \frac{4}{\sin^2 (2x)}. \end{align*}$

Then, we take the derivative,

$\begin{align*} f'(x) &= \frac{-4(2 \cos (2x))(\sin (2x))(2)}{\sin^4 (2x)} \\[8pt] &= - \frac{16 \cos (2x)}{\sin^3 (2x)}. \end{align*}$

Compute the derivative of the given function

by RoRi

September 21, 2015

Compute the derivative of the function

$f(x) = \tan \left( \frac{x}{2} \right) - \cot \left( \frac{x}{2} \right).$

We recall the derivatives $(\tan x)' = \sec^2 x$ and $(\cot x)' = \csc^2 x$ . (We proved these in this exercise, found in Apostol Section 4.6, Exercise #25.) Then, we use the chain rule to compute

$\begin{align*} f'(x) &= \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) + \frac{1}{2} \csc^2 \left( \frac{x}{2} \right) \\[8pt] &= \frac{1}{2} \left( \frac{\sin^2 \left( \frac{x}{2} \right) + \cos^2 \left( \frac{x}{2} \right)}{\cos^2 \left( \frac{x}{2} \right) \sin^2 \left(\frac{x}{2} \right)}\right) \\[8pt] &= \frac{1}{2} \left( \frac{1}{\left(\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right)\right) \left(\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \right)}\right) \\[8pt] &= \frac{1}{2} \left( \frac{1}{\left(\frac{1}{2} \sin x\right) \left(\frac{1}{2} \sin x\right)} \right) \\ &= \frac{2}{\sin^2 x}. \end{align*}$

The second to last equality follows from the trig identity $\sin (2x) = 2 \sin x \cos x$ since,

$\sin x = \sin 2\left(\frac{x}{2}\right) = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right).$

Implies,

$\sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) = \frac{1}{2} \sin x.$

Compute the derivative of the given function

by RoRi

September 20, 2015

Compute the derivative of the function

$f(x) = \frac{\sin^2 x}{\sin (x^2)}.$

Using the chain rule we compute,

$\begin{align*} f'(x) &= \frac{(2 \sin x \cos x)(\sin (x^2)) - (\sin^2 x)(2x \cos (x^2))}{\sin^2 (x^2)} \\[8pt] &= \frac{2 \sin x \big(\cos x \sin (x^2) - x \sin x \cos (x^2)\big)}{\sin^2 (x^2)} \end{align*}$

Compute the derivative of the given function

by RoRi

September 20, 2015

Compute the derivative of the function

$f(x) = \sin (\sin (\sin x)).$

Using the chain rule we compute,

$f'(x) = \cos(\sin (\sin x)) \cdot \cos(\sin x) \cdot \cos x.$

Compute the derivative of the given function

by RoRi

September 20, 2015

Compute the derivative of the function

$f(x) = \sin^n x \cdot cos (nx).$

Using the chain rule we compute,

$\begin{align*} f'(x) &= n (\sin^{n-1} x) \cos x \cos (nx) + \sin^n x (- \sin nx)(n) \\ &= (n \sin^{n-1} x)( \cos x \cos (nx) - \sin x \sin (nx)) \\ &= (n \sin^{n-1} x)\cos ((n+1)x). \end{align*}$

The final equality follows from the identity $\cos (x+y) = \cos x \cos y - \sin x \sin y$ since

$\cos ((n+1)x) = \cos (x + nx) = \cos x \cos (nx) - \sin x \sin (nx).$

Compute the derivative of the given function

by RoRi

September 20, 2015

Compute the derivative of the function

$f(x) = \sin (\cos^2 x) \cdot \cos (\sin^2 x).$

Using the chain rule we compute,

$\begin{align*} f'(x) &= \cos (\cos^2 x)(-2 \cos x \sin x)\cos (\sin^2 x) + \sin (\cos^2 x)(-\sin (\sin^2 x))2 \sin x \cos x \\ &= (-2 \sin x \cos x)(\cos (\cos^2 x)\cos (\sin^2 x) + \sin (\cos^2 x)\sin (\sin^2 x)) \\ &= -\sin (2x) (\cos (\cos^2 x - \sin^2 x)) \\ &= - \sin (2x) \cdot \cos(\cos (2x)). \end{align*}$

Compute the derivative of the given function

by RoRi

September 20, 2015

Compute the derivative of the function

$f(x) = (2-x^2)\cos (x^2) + 2x \sin (x^3).$

Using the chain rule we compute,

$\begin{align*} f'(x) &= (-2x)\cos (x^2) + (2-x^2)(2x)(-\sin (x^2)) + 2 \sin (x^3) + (2x)(3x^2) \cos (x^3) \\ &= (2x^3 -4x)\sin (x^2) - 2x \cos (x^2) + 2 \sin (x^3) + 6x^3 \cos (x^3). \end{align*}$