Calculus - Differentiation - Chain Rule

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Question 1


Use the chain rule: when $y=(ax+b)^n$, $\frac{\mathrm{d}y}{\mathrm{d}x}=an{(ax+b)}^{n-1}$ to differentiate the following:

a) $y={(2x+1)}^3$a s v d
b) $y={(3x+2)}^4$a s v d
c) $y={(8x+15)}^{10}$a s v d
d) $y={(x+3)}^{7}$a s v d
e) $y={(3x+4)}^{2}$a s v d
f) $y={(4x-6)}^6$a s v d
g) $y={(-2x+3)}^5$a s v d
h) $y={(-5x-1)}^8$a s v d
i) $y={(6x+5)}^{-2}$a s v d
j) $y={(-2x-1)}^{-3}$a s v d
k) $y={(8x+1)}^{\frac32}$a s v d
l) $y={(5x+3)}^{\frac12}$a s v d


Question 2


First write in the form $y=(ax+b)^n$ and then differentiate:

a) $y={(1+2x)}^{3}$a s v d
b) $y={(8-4x)}^{5}$a s v d
c) $y={(-3-3x)}^{-3}$a s v d
d) $y={(1+x)}^{-1}$a s v d
e) $y=\frac{1}{1+x}$a s v d
f) $y=\frac{1}{x-3}$a s v d
g) $y=\frac{1}{{(x+4)}^2}$a s v d
h) $y=\frac{1}{{(2x-3)}^3}$a s v d
i) $y=\frac{1}{{(5-2x)}^5}$a s v d
j) $y=\sqrt{(2x+1)}$a s v d
k) $y=\sqrt{{(6x+4)}^3}$a s v d
l) $y=\frac{1}{\sqrt[4]{{(16x-3)}^3}}$a s v d


Question 3


Use the chain rule and the fact that when $y=af(x)$, $\frac{\mathrm{d}y}{\mathrm{d}x}=af'(x)$ to differentiate the following:

a) $y=2{(2x+1)}^3$a s v d
b) $y=5{(3x+2)}^4$a s v d
c) $y=-2{(8x+3)}^6$a s v d
d) $y=-7{(x+3)}^5$a s v d
e) $y=-3{(-2x-4)}^6$a s v d
f) $y=5{(4-3x)}^2$a s v d
g) $y=-3{(6-x)}^{-1}$a s v d
h) $y=6{(6x-3)}^{\frac43}$a s v d
i) $y=2\sqrt[3]{(3+12x)}$a s v d


Question 4


Use the chain rule to differentiate the following:

a) $y={(x^2+1)}^3$a s v d
b) $y={(x^3+7)}^4$a s v d
c) $y={(2x^2+5)}^2$a s v d
d) $y={(3x^4-15)}^7$a s v d
e) $y={(15-3x^3)}^3$a s v d
f) $y=5{(3x^2-22)}^8$a s v d
g) $y=-5{(2x^3+15)}^5$a s v d
h) $y={(x^2+x)}^3$a s v d
i) $y={(4x^5+3x^2)}^6$a s v d
j) $y={(x^2+x+1)}^4$a s v d
k) $y={(4x^3+x^2-24)}^3$a s v d
l) $y={(3-2x^2-5x^3+6x^6)}^7$a s v d


Question 5


Differentiate these:

a) $y=\frac{2}{{(2x+5)}^3}$a s v d
b) $y=-\frac{10}{{(5-x)}^2}$a s v d
c) $y=\frac{-4}{{(2x^2+3x-1)}^6}$a s v d
d) $y=6\sqrt[5]{(3x^4-18x+2)}$a s v d
e) $y=-\frac{2}{\sqrt{{(2-x^3)}^7}}$a s v d
f) $y=-\frac{22}{\sqrt[3]{{(5-x^2+4x^5)}^{10}}}$a s v d


Question 6


Find the answer to the following worded questions:

a) Find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $y={(2x+4)}^5$ when $x=3$a s v d

b) Find the value of $\frac{\mathrm{d}y}{\mathrm{d}x}$ when $y=2{(3x-4)}^3$ when $x=2$a s v d

c) Find the gradient on the graph $y={(5-x)}^3$ when $x=1$a s v d

d) Find the gradient on the graph $y={(x^2-1)}^3$ at the point $(2,27)$a s v d