Differentiating Algebraic and Trigonometric Functions

(kf)′=kf′ for k∈R\displaystyle \bigl( k f\bigr)' = kf' \text{ for } k \in \mathbf{R}
(f±g)′=f′±g′\displaystyle \bigl( f \pm g \bigr)' = f' \pm g'
(f∘g)′=(f′∘g)(g′)\displaystyle \bigl( f \circ g \bigr)' = \bigl(f'\circ g\bigr)(g')
(fg)′=f′g+fg′\displaystyle \bigl( f g \bigr)' = f'g + fg'
(fg)′=f′g−fg′g2\displaystyle \biggl( \frac{f}{g} \biggr)' = \frac{f'g - fg'}{g^2}
ddx(xn)=nxn−1 for n≠0\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(x^n\bigr) = nx^{n-1} \text{ for } n\neq 0
ddx(sin⁡(x))=cos⁡(x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\sin(x)\bigr) = \cos(x)
ddx(sec⁡(x))=sec⁡(x)tan⁡(x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\sec(x)\bigr) = \sec(x)\tan(x)
ddx(tan⁡(x))=sec⁡2(x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\tan(x)\bigr) = \sec^2(x)
ddx(cos⁡(x))=−sin⁡(x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\cos(x)\bigr) = -\sin(x)
ddx(csc⁡(x))=−csc⁡(x)cot⁡(x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\csc(x)\bigr) = -\csc(x)\cot(x)
ddx(cot⁡(x))=−csc⁡2(x)\displaystyle \frac{\mathrm{d}}{\mathrm{d}x} \bigl(\cot(x)\bigr) = -\csc^2(x)

Explicit Differentiation

Each of the following expressions of a variable xx can be thought of as the formula for a real-valued function that’s continuous on some open subset of the real numbers. As exercise, for each of these functions, write down an explicit formula for its derivative. Use software capable of symbolic differentiation to check your formulas.

17x2−2\displaystyle 17x^2-2
x17\displaystyle x^{17}
1x17\displaystyle \frac{1}{x^{17}}
x17\displaystyle \sqrt[17]{x}
117+1x\displaystyle \frac{1}{17}+\frac{1}{x}
17xπ\displaystyle 17x^{\pi}
1.7xπ\displaystyle 1.7x^{\pi}
1π\displaystyle \frac{1}{\pi}
1.7−1x7.1\displaystyle 1.7^{-1}x^{7.1}
1.72x−7.1\displaystyle 1.7^{2}x^{-7.1}
x2x−7.1\displaystyle x^{2}x^{-7.1}
x217−7.1\displaystyle x^{2}17^{-7.1}
x3+x2+x+1+1x+1x2+1x3\displaystyle x^3+x^2+x+1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}
(x−17)2\displaystyle \bigl(x-17\bigr)^2
(x−172)\displaystyle \bigl(x-17^2\bigr)
(x−2)17\displaystyle \bigl(x-2\bigr)^{17}
(17x−2)17\displaystyle \bigl(17x-2\bigr)^{17}
(17x2−2)17\displaystyle \bigl(17x^2-2\bigr)^{17}
(17x2−2)(x17−17x+2)\displaystyle \bigl(17x^2-2\bigr)\bigl(x^{17}-17x+2\bigr)
17x2−2x17−17x+2\displaystyle \frac{17x^2-2}{x^{17}-17x+2}
11+1x\displaystyle \frac{1}{1+\frac{1}{x}}
x27\displaystyle \sqrt[7]{\sqrt[2]{\sqrt{x}}}
71sin⁡(x)+17cos⁡(x)\displaystyle 71\sin(x) + 17\cos(x)
17sin⁡(x)cos⁡(x)\displaystyle 17\sin(x)\cos(x)
sin⁡(cos⁡(x))\displaystyle \sin\bigl(\cos(x)\bigr)
17sin⁡17(x)\displaystyle 17\sin^{17}(x)
sec⁡(x17−x)\displaystyle \sec\bigl(x^{17}-x\bigr)
csc⁡(17x−1)\displaystyle \csc\bigl(\sqrt{17x-1}\bigr)
cot⁡(x)\displaystyle \sqrt{\cot(x)}
tan⁡(sin⁡(x))\displaystyle \tan\bigl(\sin(x)\bigr)
tan⁡17(sec⁡(x))\displaystyle \tan^{17}\bigl(\sec(x)\bigr)
tan⁡(2x)csc⁡(17x)+1\displaystyle \tan(2x)\csc(17x)+1
tan⁡(2x)1+csc⁡(17x)\displaystyle \frac{\tan(2x)}{1+\csc(17x)}
11+1tan⁡(17x)\displaystyle \frac{1}{1+\frac{1}{\tan(17x)}}
tan⁡(tan⁡(x))\displaystyle \tan\bigl(\tan(x)\bigr)
csc⁡(csc⁡(x))\displaystyle \csc\bigl(\csc(x)\bigr)
sec⁡(sin⁡(x)tan⁡(x))\displaystyle \sec\bigl(\sin(x)\tan(x)\bigr)
x17cot⁡17(x)\displaystyle x^{17}\cot^{17}(x)
x17cot⁡17(x)\displaystyle \frac{\sqrt{x^{17}}}{\cot^{17}(x)}
(xsec⁡(x))17\displaystyle \biggl(\frac{\sqrt{x}}{\sec(x)}\biggr)^{17}
tan⁡(cos⁡(cos⁡(x)))\displaystyle \tan\Bigl(\cos\bigl(\cos(x)\bigr)\Bigr)
sin⁡(sec⁡(sec⁡(x)))\displaystyle \sin\Bigl(\sec\bigl(\sec(x)\bigr)\Bigr)
sin⁡(sec⁡17(x)tan⁡71(x))\displaystyle \sin\bigl(\sec^{17}(x)\tan^{71}(x)\bigr)
sin⁡(sin⁡2(x)+cos⁡2(x)tan⁡2(x)+1)\displaystyle \sin\biggl(\frac{\sin^2(x)+\cos^2(x)}{\tan^2(x)+1}\biggr)

Implicit Differentiation

Each of the following equations involves an independent variable xx and a variable yy dependent on x.x. As exercise, for each of these equations, take the derivative with respect to xx implicitly. Then if possible, express the derivative of yy explicitly as a function of xx. Use software capable of symbolic differentiation to check your formulas.

xy=1\displaystyle xy = 1
(xy)2=1\displaystyle (xy)^2 = 1
(xy)17=xy+1\displaystyle (xy)^{17} = xy+1
(x+y)17=x+y+1\displaystyle (x+y)^{17} = x+y+1
(17x+71y)2=1\displaystyle (17x+71y)^2 = 1
(17xy+71y2)2=1\displaystyle \bigl(17xy+71y^2\bigr)^2 = 1
x3+y2+x=1y+1x2+1y3\displaystyle x^3+y^2+x = \frac{1}{y}+\frac{1}{x^2}+\frac{1}{y^3}
xy=x+y\displaystyle \frac{x}{y} = x+y
x2y3=x7+y\displaystyle \frac{x^2}{y^3} = x^7+y
xyy+17=x+y\displaystyle \frac{xy}{y+17} = x+y
11+xy=1\displaystyle \frac{1}{1+\frac{x}{y}} = 1
xy+2y=1\displaystyle \sqrt{xy+2y} = 1
71sin⁡(x)=17cos⁡(y)\displaystyle 71\sin(x) = 17\cos(y)
sin⁡(x)cos⁡(y)=1\displaystyle \sin(x)\cos(y) = 1
sin⁡(xcos⁡(y))=1\displaystyle \sin\bigl(x\cos(y)\bigr) = 1
17sin⁡17(y)=x\displaystyle 17\sin^{17}(y) = x
sec⁡(y17−x)=1\displaystyle \sec\bigl(y^{17}-x\bigr) = 1
cot⁡(y)=x\displaystyle \sqrt{\cot(y)} = x
tan⁡(sin⁡(xy))\displaystyle \tan\bigl(\sin(xy)\bigr)
1y+1tan⁡(17y)=x\displaystyle \frac{1}{y+\frac{1}{\tan(17y)}}=x
sin⁡(cos⁡(sec⁡(y)))=xy\displaystyle \sin\Bigl(\cos\bigl(\sec(y)\bigr)\Bigr) = xy
tan⁡(cos⁡(cos⁡(xy)))=y\displaystyle \tan\Bigl(\cos\bigl(\cos(xy)\bigr)\Bigr) = y

Now suppose each of the previous equations involved an independent variable tt (not explicitly present) and variables xx and yy each dependent on t.t. As exercise, for each of those equations equations, take the derivative with respect to tt implicitly. Use software capable of symbolic differentiation to check your formulas.