find derivative
y= 3x^5-6x^3/7secx
find dy/dx
* xcosy= 5( y+cosx)
* X^5-Y^3+8XY=466
* A RIGHT TRIANGLE HAS LEGS OF LENGTH 15M AND 7M. THE 7M LEG IS NOT CHANGING. 15M LEG IS GETTING LONGER AT RATE OF 8M PER SECOND. SO WHAT IS RATE OF CHANGE OF THE HYPOTENUSE?
need parentheses on 1st
in any case, use quotient rule. what do you get?
x cosy = 5y + 5cosx
cosy - x*siny y' = 5y' - 5sinx
y'(x siny + 5) = cosy + 5sinx
y' = (cosy+5sinx)/(x siny+5)
h^2 = x^2+y^2
2h dh/dt = 2x dx/dt + 2y dy/dt
2√274 dh/dt = 0 + 2*15*8
dh/dt = 120/√274
where did you
get 274 from
To find the derivative of the given function y = (3x^5 - 6x^3) / (7sec(x)), we can use the quotient rule. The quotient rule states that for a function u(x) / v(x), the derivative is given by:
(dy/dx) = (u'(x) * v(x) - u(x) * v'(x)) / [v(x)]^2
Applying this rule to the given function, we have:
u(x) = 3x^5 - 6x^3
v(x) = 7sec(x)
First, let's find the derivative of u(x):
u'(x) = d/dx (3x^5 - 6x^3)
= 15x^4 - 18x^2
Next, let's find the derivative of v(x):
v'(x) = d/dx (7sec(x))
= 7sec(x) * tan(x)
Now, we can substitute these values into the quotient rule formula:
(dy/dx) = (u'(x) * v(x) - u(x) * v'(x)) / [v(x)]^2
= [(15x^4 - 18x^2) * (7sec(x)) - (3x^5 - 6x^3) * (7sec(x) * tan(x))] / [7sec(x)]^2
= [105x^4sec(x) - 126x^2sec(x) - 21x^5sec(x)tan(x) + 42x^3sec(x)tan(x)] / 49sec^2(x)
Simplifying the expression, we get the derivative:
(dy/dx) = [105x^4sec(x) - 126x^2sec(x) - 21x^5sec(x)tan(x) + 42x^3sec(x)tan(x)] / 49sec^2(x)
To find the derivative of a function, you can use the rules of differentiation. In this case, we have the function:
y = (3x^5 - 6x^3) / (7sec(x))
To find dy/dx, we can differentiate each term separately using the power rule, chain rule, and quotient rule if necessary.
Let's go step by step:
Step 1: Rewrite the expression using trigonometric identities.
Since sec(x) is the reciprocal of cos(x), we can rewrite the expression as:
y = (3x^5 - 6x^3) / (7/cos(x))
Now, we can simplify the expression to:
y = (3x^5 - 6x^3) * cos(x) / 7
Step 2: Differentiate each term.
For (3x^5 - 6x^3), we can differentiate each term term by term:
d/dx (3x^5) = 15x^4
d/dx (-6x^3) = -18x^2
For cos(x), we can differentiate it using the chain rule:
d/dx (cos(x)) = -sin(x)
Step 3: Combine the derivatives.
Using the product rule, the derivative of y is:
dy/dx = (15x^4 - 18x^2) * cos(x) / 7 + (3x^5 - 6x^3) * (-sin(x)) / 7
Simplifying the expression gives:
dy/dx = (15x^4 - 18x^2) * cos(x) - (3x^5 - 6x^3) * sin(x) / 7
That is the derivative of y with respect to x.
For the second equation:
x * cos(y) = 5(y + cos(x))
To find dy/dx, we can differentiate each term using the product rule and chain rule if necessary.
Step 1: Rearrange the equation:
x * cos(y) = 5y + 5cos(x)
Step 2: Differentiate each term.
For x, we differentiate it with respect to x:
d/dx (x) = 1
For cos(y), we differentiate using the chain rule:
d/dx (cos(y)) = -sin(y) * dy/dx
For the right-hand side, both terms are constant, so their derivatives are zero.
d/dx (5y) = 0
d/dx (5cos(x)) = 0
Step 3: Combine the derivatives.
Putting it all together, we have:
cos(y) - x * sin(y) * dy/dx = 0
Simplifying gives:
dy/dx = cos(y) / (x * sin(y))
That is the derivative dy/dx of the equation.
For the third question about the rate of change of the hypotenuse:
We have a right triangle with legs of length 15m and 7m. The length of the 7m leg is not changing, while the 15m leg is getting longer at a rate of 8m per second.
Let's denote the hypotenuse as h. By the Pythagorean theorem, we know that:
h^2 = 15^2 + 7^2
Differentiating both sides of this equation with respect to time (t):
2h * (dh/dt) = 2 * 15 * (d15/dt) + 2 * 7 * (d7/dt)
Since d7/dt is zero (as the 7m leg is not changing), we have:
2h * (dh/dt) = 2 * 15 * (d15/dt)
Simplifying:
dh/dt = (15 * d15/dt) / h
Substituting the given rate of change of the 15m leg (d15/dt = 8m/s) and the value of the hypotenuse (h = sqrt(15^2 + 7^2) = sqrt(274)), we can calculate the rate of change of the hypotenuse:
dh/dt = (15 * 8) / sqrt(274)
Evaluate this expression to find the rate of change of the hypotenuse.