Differential Functions - CLEP Calculus

Okay, don't be overwhelmed. Take this chain rule one step at a time:

Step 1: Do the sine...

cos(cos(tan(sin(x))))

Step 2: Do the cosine . . .

–sin(tan(sin(x)))

Step 3: do the tangent . . . this is the simple chain rule, so diffentiate the argument as well

sec2(sin(x))cos(x)

Step 4: Multiply them together:

–cos(cos(tan(sin(x)))) * sec2(sin(x))cos(x) * sin(tan(sin(x)))

This is a simple chain rule. The derivative of the secant is secant * tangent; therefore:

f'(x) = sec(x2 + 4x) * tan(x2 + 4x) * (2x + 4)

Distribute everything to get your answer: 2x * sec(x2 + 4x)tan(x2 + 4x) + 4sec(x2 + 4x)tan(x2 + 4x)

Consider this as a chain rule case. Do each step:

Step 1: cos4

4cos3(x2)

Step 2: cos(x2); this can be treated like a normal case of the chain rule

–sin(x2) * 2x

Combining these, we get

–8x * sin(x2)cos3(x2)

f(x) = (100/x2) + (50/x) – 200x2

First, rewrite the equation: 100x–2 + 50x–1 – 200x2

At this point, it is relatively easy to differentiate:

f'(x) = –2 * 100x–3 – 50x–2 – 400x = (–200/x3) – (50/x2) – 400x

Simplify by making x3 the common denominator:

(–200 – 50x – 400x4)/x3

Factor out the common –50 in the numerator to make things look nicer:

–50(4 + x + 8x4)/x3

f(x) = 2x * ln(sin(x))

Here, we have a product rule with a chain rule.

The derivative of ln(sin(x)) = (1/sin(x)) * cos(x). Note that this is cot(x).

Therefore, our full derivative is:

f'(x) = 2ln(sin(x)) + 2x * cot(x) = 2(ln(sin(x)) + x * cot(x))

f(x) = sin4(x) – cos3(x)?

Treat these as chain rules:

The derivative of sin4(x) is therefore: 4sin3(x)cos(x)

The derivative of cos3(x) is therefore: 3cos2(x) * –sin(x) or –3cos2(x)sin(x)

This gives us 4sin3(x)cos(x) + 3cos2(x)sin(x), which really cannot be simplified any further.

This is a case of the chain rule.

Step 1: Deal with the exponential function

ln(2) * 2tan(x)

Step 2: Deal with the exponent

sec2(x)

Multiply them together to get your answer:

sec2(x) * ln(2) * 2tan(x)

Applying the chain rule to each element, we get:

f'(x) = 4sin3(x)cos(x) + 4cos3(x)sin(x)

If we factor out the common factors, we get:

f'(x) = 4sin(x)cos(x)(sin2(x) + cos2(x)) = 4sin(x)cos(x)(1) = 4sin(x)cos(x)

Also, since we know that 2sin(x)cos(x) = sin(2x), we know that 4sin(x)cos(x) = 2sin(2x)

This is a product rule combined with a chain rule. Let's do the chain rule for sin(4x3) first:

cos(4x3) * 12x2 = 12x2cos(4x3)

With this in mind, let's solve our whole problem:

2x * sin(4x3) + x2 * 12x2cos(4x3) = 2x * sin(4x3) + 12x4cos(4x3) = 2x(sin(4x3) + 6x3cos(4x3))

The first element is merely differentiated as 4x3

The second element is a relatively simple product rule:

sin(x–5) + x * cos(x–5) * –5 * x–6 = sin(x–5) + cos(x–5) * –5 * x–5 = sin(x–5) – (5 * cos(x–5)/x5)

Put everything back together:

4x3 – ( sin(x–5) – (5 * cos(x–5)/x5) ) = 4x3 – sin(x–5) + (5 * cos(x–5)/x5)

This is just a normal product rule problem:

5 * ln(2x) + 5x * 2 * (1/2x)

Simplify: 5ln(2) + 5 = 5(ln(2) + 1)

First rewrite your function to make this easier:

f(x) = 6/x2 = 6x–2

Now, we must find the first derivative:

f'(x) = –2 * 6 * x–3 = –12/x3

The slope of the tangent line of f(x) at x = 2 is: f'(2) = –12/23 = –12/8 = –3/2 = –1.5

First we must find the first derivative of f(x).

f'(x) = 4x3 + 12x–5

To find the slope of the tangent line of f(x) at 5, we merely have to evaluate f'(x) at 5:

f'(5) = 4*53 + 12* 5–5 = 500 + 12/3125 = 500.00384

This requires both the use of the chain rule and the product rule. Start with the natural logarithm: 2/(cos(x)sin(x))

Now, multiply by d/dx cos(x)sin(x), which is: –sin(x)sin(x) + cos(x)cos(x) = cos2(x) – sin2(x)

Therefore, f'(x) = (cos2(x) – sin2(x)) * 2/(cos(x)sin(x)) = 2(cos2(x) – sin2(x))sec(x)csc(x)

From our trigonometric identities, we know cos2(x) – sin2(x) = cos(2x)

Therefore, we can finilize our simplification to f'(x) = 2cos(2x)sec(x)csc(x)

We can use trigonometric identities to transform integrals that we typically don't know how to integrate.

Thus,

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

It is essentially the slope of a multi-dimensional function at any given point

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of

x:

y:

z:

The slope is

Use the quotient rule.

The derivative is .

Use the quotient rule to find the derivative.

Thus, the derivative is .

Use the product rule to find the derivative.

Simplify, and the derivative is .