Method OF Differentiation flash cards

1. 1.State the Chain Rule for differentiating a composite function $y=f(g(x))$.

   $\dfrac{dy}{dx}=f'(g(x))\cdot g'(x)$, i.e. differentiate the outer function keeping the inner function intact, then multiply by the derivative of the inner function.

   Hint: Think of peeling layers from outside in.

2. 2.How do you differentiate $y=f(x)^{g(x)}$ (variable base and variable exponent)?

   Take log of both sides: $\ln y=g(x)\ln f(x)$, then differentiate implicitly w.r.t. $x$: $\dfrac{1}{y}\dfrac{dy}{dx}=g'(x)\ln f(x)+g(x)\dfrac{f'(x)}{f(x)}$, so $\dfrac{dy}{dx}=y\left[g'(x)\ln f(x)+\dfrac{g(x)f'(x)}{f(x)}\right]=f(x)^{g(x)}\left[g'(x)\ln f(x)+\dfrac{g(x)f'(x)}{f(x)}\right]$.

   Hint: This technique is called logarithmic differentiation.

3. 3.If $x=\phi(t)$ and $y=\psi(t)$ are given parametrically, find $\dfrac{dy}{dx}$.

   $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{\psi'(t)}{\phi'(t)}$, provided $\phi'(t)\neq0$.

   Hint: Differentiate $x$ and $y$ separately with respect to the parameter $t$, then divide.

4. 4.How do you find $\dfrac{dy}{dx}$ for an implicit relation like $x^2+y^2=r^2$?

   Differentiate both sides w.r.t. $x$, treating $y$ as a function of $x$ and using the chain rule on $y$-terms: $2x+2y\dfrac{dy}{dx}=0$, giving $\dfrac{dy}{dx}=-\dfrac{x}{y}$ (for $y\neq0$).

   Hint: Apply $\dfrac{d}{dx}(y^n)=ny^{n-1}\dfrac{dy}{dx}$ to every $y$-term.

5. 5.How is the derivative of one function $u=f(x)$ with respect to another function $v=g(x)$ computed?

   $\dfrac{du}{dv}=\dfrac{du/dx}{dv/dx}=\dfrac{f'(x)}{g'(x)}$, where both $u$ and $v$ are differentiated w.r.t. the common variable $x$ and the results are divided (provided $g'(x)\neq0$).

   Hint: Useful for problems like differentiating $\sin^{-1}x$ with respect to $\cos^{-1}x$.

6. 6.If $y=x^{x^{x}}$ (a tower of three $x$'s), find $\dfrac{dy}{dx}$.

   Let $t=x^x$, so $\ln t=x\ln x\Rightarrow t'=x^x(1+\ln x)$. Now $y=x^{t}$, so $\ln y=t\ln x$, giving $\dfrac{y'}{y}=t'\ln x+\dfrac{t}{x}$. Substituting $t$ and $t'$: $y'=x^{x^{x}}\left[x^{x}(1+\ln x)\ln x+x^{x-1}\right]$.

   Hint: Take logs twice, from the top of the tower down.

7. 7.Differentiate $y=(\sin x)^{\cos x}+(\cos x)^{\sin x}$.

   For $u=(\sin x)^{\cos x}$: $\ln u=\cos x\ln\sin x$, so $u'=u\left[\dfrac{\cos^2x}{\sin x}-\sin x\ln\sin x\right]$. For $v=(\cos x)^{\sin x}$: $\ln v=\sin x\ln\cos x$, so $v'=v\left[\cos x\ln\cos x-\dfrac{\sin^2x}{\cos x}\right]$. Then $y'=u'+v'=(\sin x)^{\cos x}\left[\dfrac{\cos^2x}{\sin x}-\sin x\ln\sin x\right]+(\cos x)^{\sin x}\left[\cos x\ln\cos x-\dfrac{\sin^2x}{\cos x}\right]$.

   Hint: Log-differentiate each term separately, then add the two derivatives.

8. 8.Differentiate $y=x^{\sin x}+(\sin x)^{x}$.

   For $u=x^{\sin x}$: $\ln u=\sin x\ln x$, so $u'=x^{\sin x}\left(\cos x\ln x+\dfrac{\sin x}{x}\right)$. For $v=(\sin x)^x$: $\ln v=x\ln\sin x$, so $v'=(\sin x)^x(\ln\sin x+x\cot x)$. Hence $y'=x^{\sin x}\left(\cos x\ln x+\dfrac{\sin x}{x}\right)+(\sin x)^x(\ln\sin x+x\cot x)$.

   Hint: Two different exponent-of-function forms — log-differentiate each independently.

9. 9.Using logarithmic differentiation, find $\dfrac{dy}{dx}$ for $y=\sqrt{\dfrac{(x-1)(x-2)}{(x-3)(x-4)}}$.

   Take $\ln y=\dfrac12\big[\ln(x-1)+\ln(x-2)-\ln(x-3)-\ln(x-4)\big]$. Differentiating, $\dfrac{y'}{y}=\dfrac12\left[\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{1}{x-4}\right]$, so $y'=\dfrac{y}{2}\left[\dfrac{1}{x-1}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{1}{x-4}\right]$ with $y$ as given.

   Hint: Convert the product/quotient/root into a sum of logs first.

10. 10.Find the value of $x>0$ at which $y=x^{1/x}$ attains its maximum, using logarithmic differentiation.

    $\ln y=\dfrac{\ln x}{x}$, so $\dfrac{y'}{y}=\dfrac{1-\ln x}{x^2}$, giving $y'=x^{1/x}\cdot\dfrac{1-\ln x}{x^2}$. Since $x^{1/x}>0$ and $x^2>0$, $y'=0\iff \ln x=1\iff x=e$. As $y'>0$ for $xe$, $x=e$ is a maximum, with maximum value $y=e^{1/e}$.

    Hint: Log-differentiate, set $y'=0$, then check the sign change.

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