
What is the derivative of \[y = {\sec ^3}(x)\]
Answer
512.1k+ views
Hint: Here in this question, we have to find the derivative of the given function, here the function is a trigonometric function. To solve this, we use the standard differentiation formulas of trigonometry function. The function also contains the power function, the differentiation formula is applied first for the power function and then for the trigonometric function.
Complete step-by-step answer:
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. In trigonometry we have six trigonometry ratios namely sine, cosine, tangent, cosecant, secant and cotangent.
Now we consider the given question
\[y = {\sec ^3}(x)\]
On applying the differentiation with respect to \[x\], we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = \dfrac{d}{{dx}}({\sec ^3}(x))\]
Since the function involves the power function and the trigonometric function, we first apply the differentiation to the power function and then for the trigonometric function
The differentiation formula for the power function is given by \[\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}\], so we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = 3\,{\sec ^{3 - 1}}(x)\dfrac{d}{{dx}}(\sec (x))\]
On simplifying we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = 3\,{\sec ^2}(x)\dfrac{d}{{dx}}(\sec (x))\]
The differentiation formula for the secant trigonometry ratio is given by \[\dfrac{d}{{dx}}(\sec (x)) = \sec (x).\tan (x)\], so we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = 3\,{\sec ^2}(x)\left( {\sec (x).\tan (x)} \right)\]
On simplifying the above term we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3\,{\sec ^3}(x)\tan (x)\]
Hence determined the solution for the given question
Therefore the derivative of \[{\sec ^3}(x)\] is \[3\,{\sec ^3}(x)\tan (x)\]
So, the correct answer is “\[{\sec ^3}(x)\] is \[3\,{\sec ^3}(x)\tan (x)\]”.
Note: The student must know about the differentiation formulas for the trigonometry function and these differentiation formulas are standard. If the function is a power function of any other function the differentiation is applied first for the power function and then to the other function.
Complete step-by-step answer:
In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. In trigonometry we have six trigonometry ratios namely sine, cosine, tangent, cosecant, secant and cotangent.
Now we consider the given question
\[y = {\sec ^3}(x)\]
On applying the differentiation with respect to \[x\], we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = \dfrac{d}{{dx}}({\sec ^3}(x))\]
Since the function involves the power function and the trigonometric function, we first apply the differentiation to the power function and then for the trigonometric function
The differentiation formula for the power function is given by \[\dfrac{d}{{dx}}({x^n}) = n.{x^{n - 1}}\], so we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = 3\,{\sec ^{3 - 1}}(x)\dfrac{d}{{dx}}(\sec (x))\]
On simplifying we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = 3\,{\sec ^2}(x)\dfrac{d}{{dx}}(\sec (x))\]
The differentiation formula for the secant trigonometry ratio is given by \[\dfrac{d}{{dx}}(\sec (x)) = \sec (x).\tan (x)\], so we have
\[ \Rightarrow \dfrac{d}{{dx}}(y) = 3\,{\sec ^2}(x)\left( {\sec (x).\tan (x)} \right)\]
On simplifying the above term we have
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 3\,{\sec ^3}(x)\tan (x)\]
Hence determined the solution for the given question
Therefore the derivative of \[{\sec ^3}(x)\] is \[3\,{\sec ^3}(x)\tan (x)\]
So, the correct answer is “\[{\sec ^3}(x)\] is \[3\,{\sec ^3}(x)\tan (x)\]”.
Note: The student must know about the differentiation formulas for the trigonometry function and these differentiation formulas are standard. If the function is a power function of any other function the differentiation is applied first for the power function and then to the other function.
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