Question

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.23. y = (x²+1)sech(ln x)(Hint: Before differentiating, express in terms of exponentials and simplify.)

Accepted Answer

Recognize that the function is a product of two functions: $$y = (x$$^{2}$$ + 1) \cdot 	ext{sech}(\ln x). We $$will need to use the product rule for differentiation, which states: $$\frac{d}{dx}[u \cdot v] = u'v + uv'. $$Rewrite the hyperbolic secant function in terms of exponentials. Recall that $$	ext{sech}(z) = \frac{2}{e$$^{z}$$ + e$$^{-z}$$}. So$$, $$	ext{sech}(\ln x) = \frac{2}{e$$^{\ln x}$$ + e$$^{-\ln x}$$}. $$Simplify the expression inside the denominator using properties of logarithms and exponentials: $$e^{\ln x} = x$ and e$$^{-\ln x}$$ = \frac{1}{x}. $$Thus, $$	ext{sech}(\ln x) = \frac{2}{x + \frac{1}{x}}. $$Simplify the denominator further by combining terms: $$x + \frac{1}{x} = \frac{x$$^{2}$$ + 1}{x}. $$Therefore, $$	ext{sech}(\ln x) = \frac{2}{\frac{x$$^{2}$$ + 1}{x}} = \frac{2x}{x$$^{2}$$ + 1}. $$Rewrite the original function using this simplification: $$y = (x$$^{2}$$ + 1) \cdot \frac{2x}{x$$^{2}$$ + 1}. $$Notice that $$(x^{2} + 1$$)$$ cancels out, so y = 2x$. Now, differentiate y = 2x$ with respect to x$ using basic differentiation rules.

In Exercises 13–24, find the derivative of y with respect to the appropriate variable.
23. y = (x²+1)sech(ln x)
(Hint: Before differentiating, express in terms of exponentials and simplify.)

Risposta video verificata per un problema simile:

Concetti chiave

Hyperbolic Secant Function and Its Exponential Form

Chain Rule

Product Rule