Question

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question in picture

Answer 1

(a) Find $\dfrac{dy}{dx}$ for each of the following

$(i)\ y=\sin^{-1}(\tan(x^2))$

$${dy\over dx}={1\over \sqrt{1-\tan^2(x^2)}}({1\over \cos^2(x^2)})(2x)={2x\sec^2(x^2)\over \sqrt{1-\tan^2(x^2)}}$$

$(ii)\ x^2e^{-3y}+y^3=6$

Differentiate both sides with respect to $x$

$${d\over dx}(x^2e^{-3y}+y^3)={d\over dx}(6)$$

Use the Chain Rule

$$2xe^{-3y}-3x^2e^{-3y}\cdot{dy\over dx}+3y^2{dy\over dx}=0$$

Solve for ${dy\over dx}$

$${dy\over dx}={2xe^{-3y}\over 3(x^2e^{-3y}-y^2)}={2x\over 3(x^2-y^2e^{3y})}$$

$(iii)\ y=(2-x)^{1\over x}$

$$\ln(y)={1\over x}\ln(2-x)$$

Differentiate both sides with respect to $x.$ Use the Chain Rule

$${1\over y}\cdot{dy\over dx}=-{1\over x^2}\ln(2-x)-{1\over x}\cdot{1\over 2-x}$$

$${dy\over dx}=\bigg(-{\ln(2-x)\over x^2}-{1\over x(2-x)}\bigg)(2-x)^{1\over x}$$

$(iv)\ y=\displaystyle\int_{0}^{\sqrt{x}}\ln(t^4+1)dt$

Use the Fundamental Theorem of Calculus and Chain Rule

$${dy\over dx}=\ln((\sqrt{x})^4+1){d\over dx}(\sqrt{x})={\ln(x^2+1)\over 2\sqrt{x}}$$

(b) Let $f(x)=xe^{-2x}.$

(i) Find the relative extreme point of $f$ on $\R$

Domain: $(-\infin,\infin)$

$f'(x)=e^{-2x}-2xe^{-2x}$

Find the critical number(s)

$f'(x)=0=>e^{-2x}-2xe^{-2x}=0=>$

$=>e^{-2x}(1-2x)=0=>x={1\over 2}$

Critical number: ${1\over 2}$

$$f({1\over 2})={1\over 2}e^{-2({1\over 2})}={1\over 2e}$$

First Derivative Test

If $x<{1\over 2},\ f'(x)>0, f(x)$ increases.

If $x>{1\over 2},\ f'(x)<0, f(x)$ decreases.

The function $f(x)$ has a relative maximum with value of ${1\over 2e}$ at $x={1\over 2}.$

$Point\ \bigg(\dfrac{1}{2},\dfrac{1}{2e}\bigg)$

(ii) Use the result of (b)(i) to find the number of real roots, if any, of the equation $f(x)=c,$ where $c$ is a positive constant.

$0<c<\dfrac{1}{2e},$ two real roots,

$c=\dfrac{1}{2e},$ one real root $x=\dfrac{1}{2},$

$c>\dfrac{1}{2e},$ there is no real root.