In mathematics, the term calculus is frequently used to perform various kinds of tasks such as finding the slope of the tangent line, the area under the curve, volume, etc. It is divided into two main types such as integral and differentiation.

In integral and differential problems, advanced functions are used, so complex calculus problems are difficult to perform.  The process of finding the integral of the function is rather tricky than differentiation. In this post, we will learn how to solve advanced derivative problems along with examples. 

What is derivative in calculus?

In calculus, the rate of change of one function with respect to another function is said to be the derivative. The process of finding the derivative of the function is said to be the differential or differentiation. 

Derivatives are defined as slopes of lines lying tangent to curves at certain points. An alternate derivative definition consists of calculating the instantaneous rate of change as the time between measurements decreases to zero. 

There are many kinds of derivatives, including linear, power, polynomial, exponential, and logarithmic derivatives. These different kinds of functions can be differentiated by using different techniques. 

Types of derivative in calculus

There are various kinds of differentiation in calculus on the basis of the nature of the function such as:

1. Explicit differentiation
2. Implicit differentiation
3. Partial differentiation 
4. Directional differentiation

The above four types are the most frequently used types of derivatives in calculus. The first type is explicit differentiation and it is used to evaluate single-variable functions such as linear, algebraic, trigonometric, etc., in which only one independent variable is involved. 

It is denoted by d/dx or f’(x). the second type of derivative is implicit differentiation. It is frequently used to find the derivative of the dependent variable with respect to an independent variable. It is denoted by dy/dx or f’(y).

The third type of derivative in calculus is partial differentiation. It is used to find the derivative of the multivariable function. It is denoted by ?/?x ?y. The fourth type of differential is the directional derivative. It is used to find the direction of the function with the unit tangent vector. 

Formulas of derivatives in calculus

Here are a few formulas of derivatives in calculus.

Derivative by using the first principle

d/dy f(y) = lim_h?0 [f(y + h) – f(y)] / h

Derivative by using rules

Constant rule	d/dy [C] = 0
Power rule	d/dy yn = n yn+1
Sum rule	d/dy [f(y) + g(y)] = d/dy [f(y)] + d/dy [g(y)]
Difference Rule	d/dy [f(y) – g(y)] = d/dy [f(y)] – d/dy [g(y)]
Product Rule	d/dy [f(y) * g(y)] = d/dy [f(y)] * g(y) + f(y) d/dy [g(y)]
Quotient Rule	d/dy [f(y) * g(y)] = 1/[g(y)]2 [d/dy [f(y)] * g(y) – f(y) d/dy [g(y)]

Derivative of trigonometric functions,

1. d/dy [sin(y)] = cos(y)
2. d/dy [cos(y)] = -sin(y)
3. d/dy [tan(y)] = sec²(y)
4. d/dy [cot(y)] = -cosec²(y)
5. d/dy [sec(y)] = sec(y) tan(y)
6. d/dy [cosec(y)] = -cosec(y) cot(y)

How to evaluate the advanced derivative problems?

The formulas and rules of differentiation in calculus are helpful in solving advanced differentiation problems.

Example 1

Evaluate the derivative of the given function with respect to “y”.

f(y) = 4y³ – 3y² + 2tan(y) – 12y⁴ + sec(y) 

Solution 

Step 1: Take the given function and write it according to the general differential expression. 

f(y) = 4y³ – 3y² + 2tan(y) – 12y⁴ + sec(y) 

d/dy [f(y)] = d/dy [4y³ – 3y² + 2tan(y) – 12y⁴ + sec(y)]

Step 2: Now use the sum and difference rules of differentiation in calculus to write the notation of differential with each function separately.

= d/dy [4y³ – 3y² + 2tan(y) – 12y⁴ + sec(y)]

= d/dy [4y³] – d/dy [3y²] + d/dy [2tan(y)] – d/dy [12y⁴] + d/dy [sec(y)]

Step 3: Now use the constant function rule of differentiation.

= d/dy [4y³ – 3y² + 2tan(y) – 12y⁴ + sec(y)]

= 4d/dy [y³] – 3d/dy [y²] + 2d/dy [tan(y)] – 12d/dy [y⁴] + d/dy [sec(y)]

Step 4: Now differentiate the above expression with the help of the power rule and trigonometric formulas. 

d/dy [4y³ – 3y² + 2tan(y) – 12y⁴ + sec(y)] = 4 [3 y^3-1] – 3 [2 y^2-1] + 2 [sec²(y)] – 12 [4 y^4-1] + [sec(y) tan(y)]

= 4 [3 y²] – 3 [2 y¹] + 2 [sec²(y)] – 12 [4 y³] + [sec(y) tan(y)]

= 12 [y²] – 6 [y] + 2 [sec²(y)] – 48 [y³] + [sec(y) tan(y)]

= 12y² – 6y + 2sec²(y) – 48y³ + [sec(y) tan(y)]

= 12y² – 6y + 2sec²(y) – 48y³ + sec(y) * tan(y)

The advanced differentiation problems can be solved easily with the help of an online derivative calculator by Allmath to get rid of the above lengthy calculations. 

Example 2

Evaluate the derivative of the given function with respect to “z”.

f(z) = 2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z) 

Solution 

Step 1: Take the given function and write it according to the general differential expression. 

f(z) = 2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z) 

d/dz [f(z)] = d/dz [2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z)] 

Step 2: Now use the sum and difference rules of differentiation in calculus to write the notation of differential with each function separately.

d/dz [2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z)] = d/dz [2z⁴] + d/dz [5z² * 2z] – d/dz [2cot(z)] + d/dz [2z³] – d/dz [cosec(z)]

Step 3: Now use the constant function rule of differentiation.

d/dz [2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z)] = 2d/dz [z⁴] + 10d/dz [z² * z] – 2d/dz [cot(z)] + 2d/dz [z³] – d/dz [cosec(z)]

Step 4: Now use the product rule of differentiation to the above expression.

d/dz [2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z)] = 2d/dz [z⁴] + 10 [z d/dz [z²] + z² d/dz [z]] – 2d/dz [cot(z)] + 2d/dz [z³] – d/dz [cosec(z)]

= 2d/dz [z⁴] + 10 [z d/dz [z²]] + 10[z² d/dz [z]] – 2d/dz [cot(z)] + 2d/dz [z³] – d/dz [cosec(z)]

Step 5: Now differentiate the above expression with the help of the power rule and trigonometric formulas. 

d/dz [2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z)] = 2 [4 z^4-1] + 10 [z [2z^2-1]] + 10[z² [z^1-1]] – 2 [-cosec²(z)] + 2 [3 z^3-1] – [-cosec(z) cot(z)]

d/dz [2z⁴ + 5z² * 2z – 2cot(z) + 2z³ – cosec(z)] = 2 [4 z³] + 10 [z [2z¹]] + 10[z² [z⁰]] – 2 [-cosec²(z)] + 2 [3 z²] – [-cosec(z) cot(z)]

= 2 [4z³] + 10 [z [2z]] + 10[z² [1]] – 2 [-cosec²(z)] + 2 [3z²] – [-cosec(z) cot(z)]

= 8z³ + 10 [2z²] + 10[z²] – 2 [-cosec²(z)] + 2 [3z²] – [-cosec(z) cot(z)]

= 8z³ + 20z² + 10z² + 2cosec²(z) + 6z² + cosec(z) cot(z)

= 8z³ + 36z² + 2cosec²(z) + cosec(z) cot(z)

Conclusion

Now you can grab the basics of differentiation from this post. We have discussed the definition, formulas, rules, and advanced examples of differentiation. Once you grab the basics of this post, you will be able to solve any kind of differential problem easily.