Question

Derivative trick Here is an alternative way to evaluate higher derivatives of a function f that may save time. Suppose you can find the Taylor series for f centered at the point a without evaluating derivatives (for example, from a known series). Then f⁽ᵏ⁾(a)=k! multiplied by the coefficient of (x−a)ᵏ. Use this idea to evaluate f⁽³⁾(0) and f⁽⁴⁾(0) for the following functions. Use known series and do not evaluate derivatives.

f(x) = eᶜᵒˢ ˣ

Accepted Answer

Recall the key idea: if the Taylor series of a function $$f(x)$$ centered at $$a is $$given by $$f(x) = \sum_{n=0}^\infty c_n (x - a)^n$$, then the $$k-th $$derivative at $$a is$$ $$f^{(k)}(a) = k$$! \cdot c_k$$. Since we want to find f$$^{(3)}$$(0)$$ and $$f^{(4)}(0$$)$$, we need the Taylor series of f(x$$) = $$e^{\cos x}$$ centered at $$0$$ and identify the coefficients of $$x^3$$ and $$x^4. $$Start by recalling the Taylor series expansion of $$\cos x$$ around $$0$$: $$\cos x = 1 - \frac{x^2$}{2!} + \frac{x^4$}{4!} - \cdots. $$Substitute this series into the exponent of $$e^{\cos x} to $$get $$e^{1 - \frac{x^2}$${2} + \frac{$$x^4$$}{24} - \cdots}$$. Then rewrite this as e^1$ \cdot e$$^{-\frac{x^2}$${2} + \frac{x^4$}{24} - \cdots}. $$Next, expand $$e^{-\frac{x^2}$${2} + \frac{$$x^4$$}{24} - \cdots}$$ as a power series in x$, keeping terms up to x^4$ since higher powers won't affect the coefficients for x^3$ and x^4$. Identify the coefficients of x^3$ and x^4$ in the full expansion e$$^{\cos x}$$, then multiply each coefficient by 3$$!$$ and 4$$!$$ respectively to find f$$^{(3)}$$(0)$$ and $$f^{(4)}(0$$)$$.

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Key Concepts

Taylor Series Expansion

Relationship Between Taylor Coefficients and Derivatives

Using Known Series to Find Derivatives