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lord [1]
3 years ago
15

Find dy/dx for y= 4^lnx

Mathematics
1 answer:
Vedmedyk [2.9K]3 years ago
6 0

Answer:

\frac{ {4}^{ ln(x) } ln(4)  }{x }

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20000/10

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For the curve with equation y = <img src="https://tex.z-dn.net/?f=x%5E%7B2%7D" id="TexFormula1" title="x^{2}" alt="x^{2}" align=
bixtya [17]

Answer:

Below.

Step-by-step explanation:

a) First we find the derivative of the function, which gives us the gradient in terms of x:-

y = x^2 + 4x + 3

Gradient at x = dy/dx = 2x + 4.

When the curve cuts the x-axis, y = 0:

x^2 + 4x + 3 = 0

(x + 1)(x + 3) = 0

so x = -1 and -3  where the curve cuts the x axis.

The gradients at these point are therefore:

At x = -1 gradient = 2(-1) + 4 = 2

at x = -3 gradient = 2(-3) + 4 = -2.

b) (i) Where the tangent is parallel to x axis the gradient = 0

so 2x + 4 = 0 giving x = -2.

We now need to substitute x = -2 into the original function to find the y coordinate.

So the coordinates are (-2, (-2)^2 + 4(-2) +3))

= (-2, -1)

(ii) First find the slope of the line 6x + 3y =  7:

3y = -6x + 7

y = -2x + 7/3.

So the slope is -2.

Thus -2 = 2x + 4

2x = -6

x = -3

So the coordinates are (-3, (-3)^2 + 4(-3) + 3)

= (-3, 0).

7 0
3 years ago
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