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These topics lie under a section of differential calculus that predicts the behaviour or rate at which variables or quantities change. This type of evaluation helps determine the rate of change in the given input value. This process of evaluation is called differentiation. One more term that comes here is integration which is the reverse of the same process. Its applications are available in all fields wherever a change of rate is analysed. Derivatives are used to find the minima and maxima of the given function, and such functions are called differential equations. Complex analysis, measure theory, functional analysis, differential geometry and abstract algebra are some streams where differential analysis is done to understand the behaviour and predict their results on different input ranges.
Derivatives of Logarithms
Definition
In logarithmic differentiation, solutions for some complicated functions are evaluated with the help of logarithms. In some cases, solving the log functions is easier when compared to differentiating a function. It can also be simplified by saying that when there are multiple terms or parameters in the form of a product in a function, this type of differentiation will help turn the same function equation into a sum of different variables. This type of equation becomes much easier to differentiate. Evaluating the derivatives becomes an easy task with the help of logarithms and chain rule properties. It can be applied to all mathematical problems that are non-zero and differential.
Log Diff formula
y=f(x)=uxvx
To solve log functions, problems that can take the shape of the above equation can be easily solved.
Below mentioned formula is used to solve differentiation problems with the help of a log.
dydxxx=xx(1+lnx)
How to solve logarithmic functions?
Step 1: Take a log on both sides of the given equation.
log y=log ux vx
log y=vxlog u(x)
Step 2: Differentiate both side with respect to x,
1ydydx=vx*1ux*u’x+log ux*v'(x)
►dydx=y[vx*1ux*u’x+log ux*v’x ]
In the above formulation, f(x) and u(x) must be positive values; they are only defined as positive values.
Let’s look at some examples of solving functions through logarithmic functions.
Steps to be followed:
- To differentiate a logarithmic function, find the natural log of the equation.
- With the help of logarithmic functions, distribute the variables.
- Differentiate with the regular differentiating formulas.
- Finally, multiply the equation by the function to get the final result.
Derivatives of Exponential
Definition
To understand the derivates of exponential, first, we need to understand the functions of exponential, i.e.,
fx=ax, in which a is greater than 0.
This is also written in the form of fx=ex, where ‘e’ refers to Euler’s number, i.e., 2.718281828.
The next section will evaluate and derive the formula for exponential functions.
Derivative of the exponential function is the product of the exponential function with natural log.
f'(a)=axln a
Now taking dy/dx, on both sides, we will get:
daxdx=ax’= axln a or for e,
dexdx=ex
So, from the above derivation, we can clearly state and understand that it predicts the variable’s growth or decay rate.
Let’s look at some examples of solving functions through exponential functions.
Steps to be followed:
- To differentiate an exponential function, differentiate the equation with the basic exponential formula.
- By applying the exponential distributive formula, separate the function.
- Differentiate with the regular differentiating formulas.
- Finally, simplify the differentiated equation.
Let’s look at some of the examples:
Example 1: xa=2alog (a)
Solution: Taking natural log on both side
dxada=dda(2alog (a) )
x’a=2aln 2log a +3awln (10), is the final result.
Example 2: fy=7y+5y2
Solution: Taking natural log both side
log (fy)=(log (7y+5y2))
f’y=7yln7+10y, is the final result.
Example 3: fa=a26a
Solution: Taking natural log both side
log (fa)=log (a26a)
f’a=2x4x+x24xln 4, is the final result.
Example 4: fz=1+9zln z
Solution: Taking natural log both side
log (fz)=log (1+9zln z)
f’z=9ln z -1+9z1zz]2 =9ln z -1z-9[ln z 2], is the final result.
Example 5: fz=2ez-5z
Solution: Taking natural log both side
log (fz)=log (2ez-5)
f’z=2ez-5z ln(5), is the final result.
Conclusion
Derivatives of logarithms and exponential are explained from the above derivations, and examples for solving them are also mentioned. The basic concept for solving these derivatives remains the same. Only the functions’ variability or the number of dependencies varies through which the complexity of the derivatives is defined. Understanding basic formulae for derivation is necessary to solve the complex derivatives, which will be very beneficial to evaluate the functions without any errors.
