In routine problems requiring differentiation, we can make use of several previously established rules for differentiating various types of functions. It is important, however, to keep in mind the meaning of the derivative as the function that gives the rate of change or gradient of the original function at each point in the domain.
Specifically, if $f$f is a smoothly continuous function, then its derivative at $x$x in the domain is:
$f'\left(x\right)=\lim_{x\rightarrow h}\frac{f\left(x+h\right)-f\left(x\right)}{h}$f′(x)=limx→hf(x+h)−f(x)h
The rules that we have established for differentiation have been deduced from and are consistent with this definition.
In practice, we make great use of the facts that:
- The derivative of a sum is the sum of the separate derivatives
- The derivative of a constant multiple of a function is the constant multiple multiplied by the derivative of the function
A summary of the main rules that we have established is below:
- Product rule: If $y=uv$y=uv, then $y'=uv'+vu'$y′=uv′+vu′
- Quotient rule: If $y=\frac{u}{v}$y=uv, then $y'=\frac{vu'-uv'}{v^2}$y′=vu′−uv′v2
- Chain rule (sometimes called the function-of-a-function rule): If $y=g\left(u\right)$y=g(u), where $u=f\left(x\right)$u=f(x), then we can write $\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}y}{\mathrm{d}u}\cdot\frac{\mathrm{d}u}{\mathrm{d}x}$dydx=dydu·dudx
- In the specific case where $y=f\left(x\right)^n$y=f(x)n, then we have $\frac{\mathrm{d}y}{\mathrm{d}x}\ =\ n\ f'\left(x\right)\ f\left(x\right)^{n-1}$dydx = n f′(x) f(x)n−1
We have also previously found the derivatives of a number of special functions:
| $\frac{\mathrm{d}}{\mathrm{d}x}e^{f\left(x\right)}$ddxef(x) | $=$= | $f'\left(x\right)\ e^{f\left(x\right)}$f′(x) ef(x) |
| $\frac{\mathrm{d}}{\mathrm{d}x}a^{f\left(x\right)}$ddxaf(x) | $=$= | $\ln a\ f'\left(x\right)\ a^{f\left(x\right)}$lna f′(x) af(x) |
| $\frac{\mathrm{d}}{\mathrm{d}x}\ln\left(f\left(x\right)\right)$ddxln(f(x)) | $=$= | $\frac{f'\left(x\right)}{f\left(x\right)}$f′(x)f(x) |
| $\frac{\mathrm{d}}{\mathrm{d}x}\log_a\left(f\left(x\right)\right)$ddxloga(f(x)) | $=$= | $\frac{f'\left(x\right)}{\ln a\ f\left(x\right)}$f′(x)lna f(x) |
| $\frac{\mathrm{d}}{\mathrm{d}x}\sin\left(f\left(x\right)\right)$ddxsin(f(x)) | $=$= | $f'\left(x\right)\cos\left(f\left(x\right)\right)$f′(x)cos(f(x)) |
| $\frac{\mathrm{d}}{\mathrm{d}x}\cos\left(f\left(x\right)\right)$ddxcos(f(x)) | $=$= | $-f'\left(x\right)\sin\left(f\left(x\right)\right)$−f′(x)sin(f(x)) |
| $\frac{\mathrm{d}}{\mathrm{d}x}\tan\left(f\left(x\right)\right)$ddxtan(f(x)) | $=$= | $f'\left(x\right)\sec^2\left(f\left(x\right)\right)$f′(x)sec2(f(x)) |
Worked examples
Example 1
Find the derivative of $\sin\left(\ln x\right)$sin(lnx).
Think: This expression is a function of another function, so to differentiate it we will be able to use the chain rule. It may be helpful to think of the "inside" function $\ln$ln and the "outside" function $\sin$sin.
Do: We differentiate the outside function first, still evaluated at $\ln x$lnx, and then multiply by the derivative of the inside function.
So for $f\left(x\right)=\sin\left(\ln x\right)$f(x)=sin(lnx), we have
$f'\left(x\right)=\cos\left(\ln x\right)\left(\frac{1}{x}\right)$f′(x)=cos(lnx)(1x)
Example 2
Find the derivative of the function $f\left(x\right)=\left(x+\sin x\right)^{\frac{3}{2}}$f(x)=(x+sinx)32 and calculate the gradient of $f$f at $x=0$x=0, $x=\frac{\pi}{4}$x=π4 and $x=\frac{\pi}{2}$x=π2.
Solution: First, we differentiate the power function to obtain:
$\frac{3}{2}\left(x+\sin x\right)^{\frac{1}{2}}$32(x+sinx)12
Then, we differentiate the inside function to obtain:
$1+\cos x$1+cosx
Finally, we have:
$f'\left(x\right)=\frac{3}{2}\left(x+\sin x\right)^{\frac{1}{2}}\left(1+\cos x\right)$f′(x)=32(x+sinx)12(1+cosx)
We calculate:
$f'\left(0\right)=\frac{3}{2}\left(0+\sin0\right)^{\frac{1}{2}}\left(1+\cos0\right)=0$f′(0)=32(0+sin0)12(1+cos0)=0
Next:
$f'\left(\frac{\pi}{4}\right)=\frac{3}{2}\left(\frac{\pi}{4}+\sin\frac{\pi}{4}\right)^{\frac{1}{2}}\left(1+\cos\frac{\pi}{4}\right)=\frac{3}{2}\sqrt{\frac{\pi}{4}+\frac{1}{\sqrt{2}}}\left(1+\frac{1}{\sqrt{2}}\right)\approx3.128$f′(π4)=32(π4+sinπ4)12(1+cosπ4)=32√π4+1√2(1+1√2)≈3.128
And lastly:
$f'\left(\frac{\pi}{2}\right)=\frac{3}{2}\left(\frac{\pi}{2}+\sin\frac{\pi}{2}\right)^{\frac{1}{2}}\left(1+\cos\frac{\pi}{2}\right)=\frac{3}{2}\sqrt{\frac{\pi}{2}+1}\approx2.405$f′(π2)=32(π2+sinπ2)12(1+cosπ2)=32√π2+1≈2.405
Example 3
Find the gradient of $f\left(x\right)=e^{-x}\left(x^3-1\right)$f(x)=e−x(x3−1) at $x=1$x=1.
Solution: Using the product rule and the function-of-a-function rule, we have:
$f'\left(x\right)=(-1)e^{-x}\left(x^3-1\right)+e^{-x}\left(3x^2\right)$f′(x)=(−1)e−x(x3−1)+e−x(3x2)
This can be written more tidily as
$f'\left(x\right)=e^{-x}\left(1+3x^2-x^3\right)$f′(x)=e−x(1+3x2−x3),
and so we have
$f'\left(1\right)=e^{-1}\times3=\frac{3}{e}\approx1.1$f′(1)=e−1×3=3e≈1.1.
Practice questions
Question 1
Differentiate $y=\sin\left(x\right)e^x$y=sin(x)ex. Give your answer in factorised form.
Question 2
Find the derivative of $y=\cos\left(\ln x\right)$y=cos(lnx).
Question 3
Find the derivative of $\ln\left(\sin x\right)$ln(sinx).