We have seen many cases where a combination of rules can be applied to find a derivative. Let's now expand on the types of problems encountered and use a combination of rules across each of the sections covered thus far. To find a derivative for more complex functions, identify the types of functions in the problem - powers, polynomials, exponential and trigonometric, and then look for structure in the function that will dictate what rules need to be applied - such as product, quotient or composites of functions.
Below is a reference list of the rules we have learned so far:
And also a summary of derivatives of some special functions and properties of derivatives:
Function | Derivative |
---|---|
$kf\left(x\right)$kf(x), where $k$k is a constant | $kf'\left(x\right)$kf′(x) |
$f\left(x\right)+g\left(x\right)$f(x)+g(x) | $f'\left(x\right)+g'\left(x\right)$f′(x)+g′(x) |
$x^n$xn | $nx^{n-1}$nxn−1 |
$e^x$ex | $e^x$ex |
$e^{f\left(x\right)}$ef(x) | $f'\left(x\right)e^{f\left(x\right)}$f′(x)ef(x) |
$\sin\left(x\right)$sin(x) | $\cos\left(x\right)$cos(x) |
$\sin\left(f\left(x\right)\right)$sin(f(x)) | $f'\left(x\right)\cos\left(f\left(x\right)\right)$f′(x)cos(f(x)) |
$\cos\left(x\right)$cos(x) | $-\sin\left(x\right)$−sin(x) |
$\cos\left(f\left(x\right)\right)$cos(f(x)) | $-f'\left(x\right)\sin\left(f\left(x\right)\right)$−f′(x)sin(f(x)) |
$\tan\left(x\right)$tan(x) | $\frac{1}{\cos^2\left(x\right)}=\sec^2\left(x\right)$1cos2(x)=sec2(x) |
$\tan\left(f\left(x\right)\right)$tan(f(x)) | $\frac{f'\left(x\right)}{\cos^2\left(f\left(x\right)\right)}=f'\left(x\right)\sec^2\left(f\left(x\right)\right)$f′(x)cos2(f(x))=f′(x)sec2(f(x)) |
Differentiate $y=e^{\sin x}+x^2\cos x$y=esinx+x2cosx.
Think: We have a range of exponential, power and trigonometric functions. The first term is a composite function of a trigonometric function within an exponential function and hence we will need to apply the chain rule. The second term is a product of a power function with a trigonometric function and hence we will need to apply the product rule.
Do:
$f\left(x\right)$f(x) | $f'\left(x\right)$f′(x) |
---|---|
$\sin x$sinx | $\cos x$cosx |
Let | $u=x^2$u=x2 | then | $u'=2x$u′=2x |
and | $v=\cos x$v=cosx | then | $v'=-\sin x$v′=−sinx |
Hence,
$\frac{dy}{dx}$dydx | $=$= | $f'\left(x\right)e^{f\left(x\right)}+uv'+vu'$f′(x)ef(x)+uv′+vu′ | |
$=$= | $\cos xe^{\sin x}+x^2\left(-\sin x\right)+\left(\cos x\right)\left(2x\right)$cosxesinx+x2(−sinx)+(cosx)(2x) |
Make appropriate substitutions |
|
$=$= | $\cos xe^{\sin x}-x^2\sin x+2x\cos x$cosxesinx−x2sinx+2xcosx |
Differentiate $y=\sin\left(x\right)e^x$y=sin(x)ex. Give your answer in factorised form.
Consider the expression $\frac{4x^2+e^x}{\cos7x}$4x2+excos7x.
By letting $u=4x^2+e^x$u=4x2+ex, find $u'$u′.
By letting $v=\cos\left(7x\right)$v=cos(7x), find $v'$v′.
Hence, find the derivative of $\frac{4x^2+e^x}{\cos\left(7x\right)}$4x2+excos(7x).
Find the equation of the tangent to the curve $y=e^{\cos x}$y=ecosx at the point $x=\frac{3\pi}{2}$x=3π2.