Derivatives & Integrals

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GRE Quantitative Reasoning › Derivatives & Integrals

Questions 1 - 10

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

Despite being a 5th degree polynomial recall that the Maclaurin series for any polynomial is just the polynomial itself, so this function's Taylor series is identical to itself with two non-zero terms.

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

For which of the following functions can the Maclaurin series representation be expressed in four or fewer non-zero terms?

$f(x)= \frac{x^{5}}{3}+2$

$f(x)= x^{2}+\sqrt{x}+1$

$f(x)=x+3\sin(x)$

$f(x)=\ln|x+3|$

Explanation

Recall the Maclaurin series formula:

$f(x)=\sum_{n=0}^{\infty}\frac{f^n(0)}{n!}x^n$

$=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+...$

The only function that has four or fewer terms is $f(x)= \frac{x^{5}}{3}+2$ as its Maclaurin series is $2+\frac{1}{3}x^5$ .

Suppose the function $f=bcx+\frac{k}{c}+b^2ln(c)$ . Solve for $\frac{\partial f}{\partial c}$ .

$bx-\frac{k}{c^2}+\frac{b^2}{c}$

$b+c+\frac{c-k}{c^2}+\frac{b^2}{c}+2b\cdot ln(c)$

$c+\frac{1}{c}+ln(c)$

$bx+b-\frac{k}{c^2}+\frac{b^2}{c}+2b\cdot ln(c)$

$bx+\frac{k}{c^2}+\frac{b^2}{c}$

Explanation

Identify all the constants in function $f=bcx+\frac{k}{c}+b^2ln(c)$ .

Since we are solving for the partial differentiation of variable , all the other variables are constants. Solve each term by differentiation rules.

$\frac{\partial f}{\partial c}=bx-\frac{k}{c^2}+\frac{b^2}{c}$

Suppose the function $f=bcx+\frac{k}{c}+b^2ln(c)$ . Solve for $\frac{\partial f}{\partial c}$ .

$bx-\frac{k}{c^2}+\frac{b^2}{c}$

$b+c+\frac{c-k}{c^2}+\frac{b^2}{c}+2b\cdot ln(c)$

$c+\frac{1}{c}+ln(c)$

$bx+b-\frac{k}{c^2}+\frac{b^2}{c}+2b\cdot ln(c)$

$bx+\frac{k}{c^2}+\frac{b^2}{c}$

Explanation

Identify all the constants in function $f=bcx+\frac{k}{c}+b^2ln(c)$ .

Since we are solving for the partial differentiation of variable , all the other variables are constants. Solve each term by differentiation rules.

$\frac{\partial f}{\partial c}=bx-\frac{k}{c^2}+\frac{b^2}{c}$

Suppose the function $f=bcx+\frac{k}{c}+b^2ln(c)$ . Solve for $\frac{\partial f}{\partial c}$ .

$bx-\frac{k}{c^2}+\frac{b^2}{c}$

$b+c+\frac{c-k}{c^2}+\frac{b^2}{c}+2b\cdot ln(c)$

$c+\frac{1}{c}+ln(c)$

$bx+b-\frac{k}{c^2}+\frac{b^2}{c}+2b\cdot ln(c)$

$bx+\frac{k}{c^2}+\frac{b^2}{c}$

Explanation

Identify all the constants in function $f=bcx+\frac{k}{c}+b^2ln(c)$ .

Since we are solving for the partial differentiation of variable , all the other variables are constants. Solve each term by differentiation rules.

$\frac{\partial f}{\partial c}=bx-\frac{k}{c^2}+\frac{b^2}{c}$

Solve for $\frac{\partial }{\partial x}$ :

Explanation

To solve for the partial derivative, let all other variables be constants besides the variable that is derived with respect to.

In , the terms are constants.

Derive as accordingly by the differentiation rules.

$\frac{\partial }{\partial x}=s+2s^2x+sy$

Differentiate the following with respect to .

$\cos x +xy^2-\sin y=y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{(2xy-\cos y -2y) }$

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{-(\sin x+y^2)}{2xy-\cos y -2y }$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{xy-\cos y -y }$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\cos x-y^2}{(2xy-\sin y -2y) }$

Explanation

Our first step is to differentiate both sides with respect to :

$\cos x \frac{\mathrm{d} }{\mathrm{d} x} +xy^2\ \frac{\mathrm{d} }{\mathrm{d} x}-\sin y \ \frac{\mathrm{d} }{\mathrm{d} x}=y^2 \ \frac{\mathrm{d} }{\mathrm{d} x}$

Note: we can differentiate the terms that are functions of with respect to , just remember to multiply it by $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$-\sin x +y^2 +2xy \ \frac{\mathrm{d} y}{\mathrm{d} x}-\cos y \ \frac{\mathrm{d} y}{\mathrm{d} x}=2y \ \frac{\mathrm{d}y }{\mathrm{d} x}$

Note: The product rule was applied above:

$xy^2 \ \frac{\mathrm{d} }{\mathrm{d} x}=x\frac{\mathrm{d} }{\mathrm{d} x}y^2+y^2\frac{\mathrm{d} y}{\mathrm{d} x}x=y^2+2yx\frac{\mathrm{d} y}{\mathrm{d} x}$

$2xy \ \frac{\mathrm{d} y}{\mathrm{d} x}-\cos y \ \frac{\mathrm{d} y}{\mathrm{d} x}-2y \ \frac{\mathrm{d}y }{\mathrm{d} x} =\sin x-y^2$

$(2xy-\cos y -2y) \ \frac{\mathrm{d}y }{\mathrm{d} x} =\sin x-y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{(2xy-\cos y -2y) }$

Differentiate the following with respect to .

$\cos x +xy^2-\sin y=y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{(2xy-\cos y -2y) }$

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{-(\sin x+y^2)}{2xy-\cos y -2y }$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{xy-\cos y -y }$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\cos x-y^2}{(2xy-\sin y -2y) }$

Explanation

Our first step is to differentiate both sides with respect to :

$\cos x \frac{\mathrm{d} }{\mathrm{d} x} +xy^2\ \frac{\mathrm{d} }{\mathrm{d} x}-\sin y \ \frac{\mathrm{d} }{\mathrm{d} x}=y^2 \ \frac{\mathrm{d} }{\mathrm{d} x}$

Note: we can differentiate the terms that are functions of with respect to , just remember to multiply it by $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

$-\sin x +y^2 +2xy \ \frac{\mathrm{d} y}{\mathrm{d} x}-\cos y \ \frac{\mathrm{d} y}{\mathrm{d} x}=2y \ \frac{\mathrm{d}y }{\mathrm{d} x}$

Note: The product rule was applied above:

$xy^2 \ \frac{\mathrm{d} }{\mathrm{d} x}=x\frac{\mathrm{d} }{\mathrm{d} x}y^2+y^2\frac{\mathrm{d} y}{\mathrm{d} x}x=y^2+2yx\frac{\mathrm{d} y}{\mathrm{d} x}$

$2xy \ \frac{\mathrm{d} y}{\mathrm{d} x}-\cos y \ \frac{\mathrm{d} y}{\mathrm{d} x}-2y \ \frac{\mathrm{d}y }{\mathrm{d} x} =\sin x-y^2$

$(2xy-\cos y -2y) \ \frac{\mathrm{d}y }{\mathrm{d} x} =\sin x-y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{\sin x-y^2}{(2xy-\cos y -2y) }$

Solve for $\frac{\partial }{\partial x}$ :

Explanation

To solve for the partial derivative, let all other variables be constants besides the variable that is derived with respect to.

In , the terms are constants.

Derive as accordingly by the differentiation rules.

$\frac{\partial }{\partial x}=s+2s^2x+sy$

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