To find the derivative, we must use both the product and the chain rule.  Let us consider each part:

f(x) = sin(x²)

g(x) = 3x² + 5x

For f(x), we will need to apply the chain rule.  Consider f(x) as being f(h(x)) where h(x) = x^2; therefore f'(x) = f'(h(x)) * h'(x) or cos(x²) * 2x = 2x * cos(x²).

Gathering together all of our data, we have:

f(x) = sin(x²)

f'(x) = 2x * cos(x²)

g(x) = 3x² + 5x

g'(x) = 6x + 5

The product rule states that if a(x) = f(x) * g(x), then a'(x) = f'(x) * g(x) + f(x) * g'(x)

For our data, then, the derivative of a(x) = sin(x²)(3x² + 5x) is:

a'(x) = 2x * cos(x²) * (3x² + 5x) + sin(x²) * (6x + 5)

Distributing, this gives us:

a'(x) = 6x³cos(x²) + 10x²cos(x²) + 6x * sin(x²)+ 5sin(x²)

For each of the parts of f(x), you will have to apply the chain rule:

For sin(2x² + 5x): cos(2x² + 5x) * (4x + 5)

For (sin(cos(x)): cos(cos(x))*(-sin(x))

Therefore, for our original:

f'(x) = cos(2x² + 5x) * (4x + 5) – cos(cos(x)) * (–sin(x))

Simplifying:

f'(x) = 4x*cos(2x² + 5x) + 5cos(2x² + 5x) + cos(cos(x)) * sin(x)

f(x) = ln(x^{2 *} tan(x²))?

The chain rule is necessary.  For the natural logarithm "portion", we know the derivative will be:1/(x² * tan(x²))

However, we must add the derivative of the argument using the product rule: 2x*tan(x²) + x² * 2x * sec²(x²)

The whole derivative: f'(x) = (2x * tan(x²) + x² * 2x * sec²(x²))/(x² * tan(x²))

Splitting apart the fraction, we have the sum of:

(2x * tan(x²))/(x² * tan(x²))

and

x² * 2x * sec²(x²)/(x² * tan(x²))

The first simplifies to 2/x

The second is a bit trickier. First let us simplify the non-trigonometric components:

2x * sec²(x²)/tan(x²)

Now, since sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x), we can rewrite:

2x * sec²(x²)/tan(x²) = 2x/(cos²(x²) * sin(x²)/cos(x²))

This reduces to: 2x/(cos(x²) * sin(x²)), which is the same as 2x * sec(x²) * csc(x²)

Therefore, we can write our solution as f'(x) = (2/x) + 2x * sec(x²) * csc(x²)

We must carefully parse this function in order to apply the chain rule correctly. Let us make the following substitutions:

f(x) = 2^x

g(x) = sin(sin(x))

s(x) = f(g(x)) = 2^sin(sin(x))

According to the chain rule, we know that s'(x)  = f'(g(x)) * g'(x)

f'(g(x)) = ln(2) * 2^sin(sin(x))

g'(x) = cos(sin(x)) * cos(x)

Therefore, s'(x) = ln(2) * 2^sin(sin(x))  * cos(sin(x)) * cos(x)

Consider each of the elements in isolation:

sin²(x) can be differentiated using the chain rule.

Step 1: 2(sin(x))

Step 2: take the derivative of sin(x): cos(x)

Therefore, d/dx (sin²(x)) = 2sin(x)cos(x)

sin(x)cos(x) can be differentiated by the product rule:

cos(x)cos(x) + sin(x) * –sin(x) = cos²(x) – sin²(x)

Combine these:

2sin(x)cos(x) – (cos²(x) – sin²(x)) = 2sin(x)cos(x) – cos²(x) + sin²(x)

This is a mixture of the product rule and the chain rule:

The first term of the product rule is: cos(x)ln(cos(x))

The second term will have sin(x) but will include the differentiation of the ln(cos(x)), which will require the chain rule:

sin(x) * (1/cos(x)) * (–sin(x)) = –sin²(x)/cos(x)

Combining both we get:

cos(x)ln(cos(x)) – sin²(x)/cos(x)

Now, note that none of the answers are the same as this ;however, we can make an alteration:

sin(x)/cos(x) is the same as tan(x)

Therefore, the answer is: cos(x)ln(cos(x)) – tan(x)sin(x)

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