FInd the differentiation $\dfrac{d}{{dx}}\log (\sec x + \tan x) = $
$1)\cos ecx$
$2)\sec x$
$3)\tan x$
$4)\cos x$
Answer
573.3k+ views
Hint: First, we need to analyze the given information which is in the trigonometric form.
> We can equate the given expression into some form and then we can differentiate using the derivatives of basic functions and applying the chain rule of differentiation.
> The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Formula used:
Chain rule of differentiation $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$
Complete step-by-step solution:
Since from the given, we have a function $\log (\sec x + \tan x)$ and we need to find its derivative part.
We know that the logarithm derivative function can be represented as $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
Now applying the chain rule, we have, $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$and apply this in the given function then we get $\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times \dfrac{d}{{dx}}(\sec x + \tan x)$ by using the logarithm derivative.
Now we can open the brackets and derivative each function, we get \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times (\dfrac{d}{{dx}}\sec x + \dfrac{d}{{dx}}\tan x)\]
We know that $\dfrac{d}{{dx}}\sec x = \sec x\tan x,\dfrac{d}{{dx}}\tan x = {\sec ^2}x$
Substituting these values, \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}}(\sec x\tan x + {\sec ^2}x)\]
Taking sec common we have, \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times \sec x(\tan x + \sec x)\]
Hence canceling the common terms, we get \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times \sec x(\tan x + \sec x) \Rightarrow \sec x\]
Therefore, the option $2)\sec x$ is correct.
Note: The main concept used in the given problem is the chain rule. We must know the derivatives of the basic functions like tangent and sec.
We also use simple algebra to simplify the expression that we will get after derivation.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $ and $\tan = \dfrac{1}{{\cot }}$
Differentiation can be defined as the derivative of the independent variable value and can be used to calculate features in an independence variance per unit modification.
In differentiation, the derivative of $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
Differentiation and integration are inverse processes like a derivative of \[\dfrac{{d({x^2})}}{{dx}} = 2x\] and the integration is $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$.
> We can equate the given expression into some form and then we can differentiate using the derivatives of basic functions and applying the chain rule of differentiation.
> The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
Formula used:
Chain rule of differentiation $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$
Complete step-by-step solution:
Since from the given, we have a function $\log (\sec x + \tan x)$ and we need to find its derivative part.
We know that the logarithm derivative function can be represented as $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
Now applying the chain rule, we have, $\dfrac{d}{{dx}}(f(g(x)) = {f^1}(g(x)) \times {g^1}(x)$and apply this in the given function then we get $\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times \dfrac{d}{{dx}}(\sec x + \tan x)$ by using the logarithm derivative.
Now we can open the brackets and derivative each function, we get \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times (\dfrac{d}{{dx}}\sec x + \dfrac{d}{{dx}}\tan x)\]
We know that $\dfrac{d}{{dx}}\sec x = \sec x\tan x,\dfrac{d}{{dx}}\tan x = {\sec ^2}x$
Substituting these values, \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}}(\sec x\tan x + {\sec ^2}x)\]
Taking sec common we have, \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times \sec x(\tan x + \sec x)\]
Hence canceling the common terms, we get \[\dfrac{d}{{dx}}\log (\sec x + \tan x) = \dfrac{1}{{\sec x + \tan x}} \times \sec x(\tan x + \sec x) \Rightarrow \sec x\]
Therefore, the option $2)\sec x$ is correct.
Note: The main concept used in the given problem is the chain rule. We must know the derivatives of the basic functions like tangent and sec.
We also use simple algebra to simplify the expression that we will get after derivation.
In total there are six trigonometric values which are sine, cos, tan, sec, cosec, cot while all the values have been relation like $\dfrac{{\sin }}{{\cos }} = \tan $ and $\tan = \dfrac{1}{{\cot }}$
Differentiation can be defined as the derivative of the independent variable value and can be used to calculate features in an independence variance per unit modification.
In differentiation, the derivative of $x$ raised to the power is denoted by $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$ .
Differentiation and integration are inverse processes like a derivative of \[\dfrac{{d({x^2})}}{{dx}} = 2x\] and the integration is $\int {2xdx = \dfrac{{2{x^2}}}{2}} \Rightarrow {x^2}$.
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