Maths Topic 7: Introductory Calculus Quiz

Answer: m=(f(x)-f(a))/(x-a)

The secant to a curve can be defined as the straight line passing through two points on the curve. Therefore, we can work out the gradient in the same way we work out gradient of any straight line, by using the rule m=(y2-y1)/(x2-x1).

Answer: f'(x)=lim h->0 (f(x+h)-f(x))/h

To solve this, we can consider two points on a curve, (x,f(x)) and (x+h,f(x+h)). Since the derivative can be used to find the gradient of a tangent, we can use the two points to show that f'(x)=lim h->0 (f(x+h)-f(x))/h.

Answer: 12

To solve this question, we simply need to differentiate the curve.
So,
y=3x^2
dy/dx=6x
Now at x=2
dy/dx=6(2)
=12

Answer: f'(x)=2x-4

To differentiate a polynomial function, we simply differentiate each term separately
So,
f(x)=x^2-4x-3
f'(x)=2x-4

Answer: 1/(2sqrt(x))

To differentiate this function, we first need to re-write it in index form
So
sqrt(x)=x^(1/2)
Now differentiating this gives,
f'(x)=(1/2)x^(-1/2)
f'(x)=(1/2)(1/x^(1/2))
f'(x)=1/(2sqrt(x))

Answer: f'(x)=10(2x+7)^4

To differentiate this function, we need to use the rule "if y=(f(x))^n, then dy/dx = n(f(x))^(n-1) * f'(x)"
So if f(x)=(2x+7)^5
Then,
f'(x)=5(2x+7)^4 * 2
f'(x)=10(2x+7)^4

Answer: f'(x)=(10x^4) ((5x+7)^2) (8x+7)

The product rule states that "if y=f(x)g(x), then dy/dx=f'(x)g(x)+g'(x)f(x)"
Now if we apply this to the given function,
f'(x)=(10x^4(5x+7)^3)+(6x^5(5x+7)^2*5)
f'(x)=(10x^4(5x+7)^3)+(30x^5(5x+7)^2)
Now simplifying this,
f'(x)=(10x^4)((5x+7)^2)((5x+7)+3x)
f'(x)=(10x^4)((5x+7)^2)(8x+7)

Answer: f'(x)=31/(5x+2)^2

The quotient rule states that
"if y=f(x)/g(x), then dy/dx=(f'(x)g(x)-g'(x)f(x))/(g(x))^2
Now if we apply this to the given function
f'(x)=(3(5x+2)-5(3x-5))/(5x+2)^2
f'(x)=(15x+6-15x+25)/(5x+2)^2
f'(x)=31/(5x+2)^2

Answer: 6x-y-11=0

To solve this question, we first need to differentiate the curve to find the gradient
So,
y=2x^3-7
dy/dx=6x^2
at x=1
dy/dx=6
Now we need to find our y-coordinate
So, at x=1
y=2(1)^3-7
y=-5
Now using the rule y-y1=m(x-x1)
y+5=6(x-1)
y+5=6x-6
6x-y-11=0
Therefore, the equation of the tangent is 6x-y-11=0.

Answer: 216x-102y+199=0

To solve this question, we first need to differentiate the curve to find the gradient.
So,
y=(2x+3)/(5x-1)
dy/dx=(2(5x-1)-5(2x+3))/(5x-1)^2
dy/dx=(10x-2-10x-15)/(5x-1)^2
dy/dx=-17/(5x-1)^2
at x=-1
dy/dx=-17/(5(-1)-1)^2
dy/dx=-17/36
Since this is the gradient of the tangent, we need to take the reciprocal to find the gradient of the normal
Therefore, dy/dx=36/17
Now using the rule y-y1=m(x-x1)
y+(1/6)=(36/17)(x+1)
17y+(17/6)=36(x+1)
17y+(17/6)=36x+36
102y+17=216x+216
216x-102y+199=0
Therefore, the equation of the normal is 216x-102y+199=0.