Differentiate with respect to x. y = (cosx)2

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11. Differentiate with respect to $x$ . $y = (\cos x)^{2x}$

Take the natural logarithm of both sides.
$\ln y = \ln ((\cos x)^{2x})$ $\ln y = 2x \ln(\cos x)$
Differentiate both sides with respect to $x$ using implicit differentiation and the product rule.
$\frac{1}{y} \frac{dy}{dx} = 2 \ln(\cos x) + 2x \left( \frac{1}{\cos x} \cdot (-\sin x) \right)$ $\frac{1}{y} \frac{dy}{dx} = 2 \ln(\cos x) - 2x \frac{\sin x}{\cos x}$ $\frac{1}{y} \frac{dy}{dx} = 2 \ln(\cos x) - 2x \tan x$
Solve for $\frac{dy}{dx}$ by multiplying by $y$ .
$\frac{dy}{dx} = y (2 \ln(\cos x) - 2x \tan x)$
Substitute $y = (\cos x)^{2x}$ back into the equation.
$\frac{dy}{dx} = (\cos x)^{2x} (2 \ln(\cos x) - 2x \tan x)$ $\frac{dy}{dx} = 2 (\cos x)^{2x} (\ln(\cos x) - x \tan x)$
The derivative is:
$\boxed{\frac{dy}{dx} = 2 (\cos x)^{2x} (\ln(\cos x) - x \tan x)}$
12. A curve passes through the point $(1,7)$ and the gradient of the curve at the point $(x,y)$ is given by $(5-2x)$ . Find the equation of the curve.
Set up the differential equation.
The gradient is given by $\frac{dy}{dx}$ .
$\frac{dy}{dx} = 5 - 2x$
Integrate both sides with respect to $x$ to find the equation of the curve.
$y = \int (5 - 2x) \, dx$ $y = 5x - 2 \frac{x^2}{2} + C$ $y = 5x - x^2 + C$
Use the given point $(1,7)$ to find the constant of integration $C$ .
Substitute $x=1$ and $y=7$ into the equation.
$7 = 5(1) - (1)^2 + C$ $7 = 5 - 1 + C$ $7 = 4 + C$ $C = 7 - 4$ $C = 3$
Write the final equation of the curve.
$\boxed{y = 5x - x^2 + 3}$
13. Solve the equation $z^2 = 5 - 12i$ .
Let $z = x + iy$ , where $x$ and $y$ are real numbers.
$(x + iy)^2 = 5 - 12i$ $x^2 + 2ixy + (iy)^2 = 5 - 12i$ $x^2 + 2ixy - y^2 = 5 - 12i$ $(x^2 - y^2) + i(2xy) = 5 - 12i$
Equate the real and imaginary parts.
$x^2 - y^2 = 5 \quad (1)$ $2xy = -12 \quad (2)$
From equation (2), express $y$ in terms of $x$ .
$y = \frac{-12}{2x} = \frac{-6}{x}$
Substitute $y$ into equation (1).
$x^2 - \left(\frac{-6}{x}\right)^2 = 5$ $x^2 - \frac{36}{x^2} = 5$
Multiply by $x^2$ to clear the denominator.
$x^4 - 36 = 5x^2$
Rearrange into a quadratic form for $x^2$ .
$x^4 - 5x^2 - 36 = 0$
Let $u = x^2$ .
Solve the quadratic equation $u^2 - 5u - 36 = 0$ .
Using the quadratic formula or factoring:
$(u - 9)(u + 4) = 0$
So, $u = 9$ or $u = -4$ .
Since $u = x^2$ and $x$ is a real number, $x^2$ must be non-negative.
Therefore, $x^2 = 9$ .
$x = \pm 3$
Find the corresponding values for $y$ .
If $x = 3$ , then $y = \frac{-6}{3} = -2$ . So, $z_1 = 3 - 2i$ .
If $x = -3$ , then $y = \frac{-6}{-3} = 2$ . So, $z_2 = -3 + 2i$ .

The solutions are:
$\boxed{z = 3 - 2i \quad \text{or} \quad z = -3 + 2i}$
14. Evaluate $\int_{0}^{2} (1 - |x - 1|)^3 dx$ .
Split the integral based on the definition of the absolute value.
The expression $|x-1|$ changes definition at $x=1$ .
For $0 \le x < 1$ : $|x-1| = -(x-1) = 1-x$ .
For $1 \le x \le 2$ : $|x-1| = x-1$ .
Rewrite the integral as a sum of two integrals.
$\int_{0}^{2} (1 - |x - 1|)^3 dx = \int_{0}^{1} (1 - (1-x))^3 dx + \int_{1}^{2} (1 - (x-1))^3 dx$ $= \int_{0}^{1} (1 - 1 + x)^3 dx + \int_{1}^{2} (1 - x + 1)^3 dx$ $= \int_{0}^{1} x^3 dx + \int_{1}^{2} (2 - x)^3 dx$
Evaluate the first integral.
$\int_{0}^{1} x^3 dx = \left[ \frac{x^4}{4} \right]_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$
Evaluate the second integral.
Let $u = 2-x$ , so $du = -dx$ .
When $x=1$ , $u=2-1=1$ .
When $x=2$ , $u=2-2=0$ .
$\int_{1}^{2} (2 - x)^3 dx = \int_{1}^{0} u^3 (-du) = - \int_{1}^{0} u^3 du$ $= \int_{0}^{1} u^3 du = \left[ \frac{u^4}{4} \right]_{0}^{1} = \frac{1^4}{4} - \frac{0^4}{4} = \frac{1}{4}$
Add the results of the two integrals.
$\int_{0}^{2} (1 - |x - 1|)^3 dx = \frac{1}{4} + \frac{1}{4} = \frac{2}{4} = \frac{1}{2}$
The value of the integral is:
$\boxed{\frac{1}{2}}$
15. Use the substitution $t = \tan \frac{\theta}{2}$ to evaluate $\int_{0}^{\frac{\pi}{2}} \frac{4}{3 + 5 \sin \theta} d\theta$ .
Apply the Weierstrass substitution formulas.
Given $t = \tan \frac{\theta}{2}$ .
Then $d\theta = \frac{2}{1+t^2} dt$ .
And $\sin \theta = \frac{2t}{1+t^2}$ .
Change the limits of integration.
When $\theta = 0$ , $t = \tan \frac{0}{2} = \tan 0 = 0$ .
When $\theta = \frac{\pi}{2}$ , $t = \tan \frac{\pi/2}{2} = \tan \frac{\pi}{4} = 1$ .
Substitute these into the integral.
$\int_{0}^{\frac{\pi}{2}} \frac{4}{3 + 5 \sin \theta} d\theta = \int_{0}^{1} \frac{4}{3 + 5 \left(\frac{2t}{1+t^2}\right)} \left(\frac{2}{1+t^2}\right) dt$ $= \int_{0}^{1} \frac{4}{\frac{3(1+t^2) + 10t}{1+t^2}} \cdot \frac{2}{1+t^2} dt$ $= \int_{0}^{1} \frac{4(1+t^2)}{3+3t^2 + 10t} \cdot \frac{2}{1+t^2} dt$ $= \int_{0}^{1} \frac{8}{3t^2 + 10t + 3} dt$
Factor the denominator.
$3t^2 + 10t + 3 = (3t+1)(t+3)$
Use partial fraction decomposition.
$\frac{8}{(3t+1)(t+3)} = \frac{A}{3t+1} + \frac{B}{t+3}$ $8 = A(t+3) + B(3t+1)$
Set $t = -3$ : $8 = A(-3+3) + B(3(-3)+1) \implies 8 = B(-8) \implies B = -1$ .
Set $t = -\frac{1}{3}$ : $8 = A(-\frac{1}{3}+3) + B(3(-\frac{1}{3})+1) \implies 8 = A(\frac{8}{3}) \implies A = 3$ .
So, the integrand becomes:
$\frac{3}{3t+1} - \frac{1}{t+3}$
Evaluate the integral.
$\int_{0}^{1} \left( \frac{3}{3t+1} - \frac{1}{t+3} \right) dt$ $= \left[ \frac{3 \ln|3t+1|}{3} - \ln|t+3| \right]_{0}^{1}$ $= \left[ \ln|3t+1| - \ln|t+3| \right]_{0}^{1}$ $= \left[ \ln\left|\frac{3t+1}{t+3}\right| \right]_{0}^{1}$
Now, apply the limits:
$= \ln\left(\frac{3(1)+1}{1+3}\right) - \ln\left(\frac{3(0)+1}{0+3}\right)$ $= \ln\left(\frac{4}{4}\right) - \ln\left(\frac{1}{3}\right)$ $= \ln(1) - \ln\left(\frac{1}{3}\right)$ $= 0 - (-\ln 3)$ $= \ln 3$
The value of the integral is:
$\boxed{\ln 3}$

Point	x	y
Minimum	−4.71	0
Maximum	−3.14	1
Minimum	−1.57	0
Maximum	0	1
Minimum	1.57	0
Maximum	3.14	1
Minimum	4.71	0

Differentiate with respect to x. y = (cosx)2

Step-by-step solution

Take the natural logarithm of both sides.

Differentiate both sides with respect to $x$ using implicit differentiation and the product rule.

Solve for $\frac{dy}{dx}$ by multiplying by $y$ .

Set up the differential equation.

Integrate both sides with respect to $x$ to find the equation of the curve.

Use the given point $(1,7)$ to find the constant of integration $C$ .

Write the final equation of the curve.

Let $z = x + iy$ , where $x$ and $y$ are real numbers.

Equate the real and imaginary parts.

From equation (2), express $y$ in terms of $x$ .

Substitute $y$ into equation (1).

Let $u = x^2$ .

Find the corresponding values for $y$ .

Split the integral based on the definition of the absolute value.

Rewrite the integral as a sum of two integrals.

Evaluate the first integral.

Evaluate the second integral.

Add the results of the two integrals.

Apply the Weierstrass substitution formulas.

Change the limits of integration.

Substitute these into the integral.

Factor the denominator.

Use partial fraction decomposition.

Evaluate the integral.

Still stuck on this one?

More Mathematics Questions

Differentiate with respect to x. y = (cosx)2

Step-by-step solution

Take the natural logarithm of both sides.

Differentiate both sides with respect to xxx using implicit differentiation and the product rule.

Solve for dydx\frac{dy}{dx}dxdy​ by multiplying by yyy.

Set up the differential equation.

Integrate both sides with respect to xxx to find the equation of the curve.

Use the given point (1,7)(1,7)(1,7) to find the constant of integration CCC.

Write the final equation of the curve.

Let z=x+iyz = x + iyz=x+iy, where xxx and yyy are real numbers.

Equate the real and imaginary parts.

From equation (2), express yyy in terms of xxx.

Substitute yyy into equation (1).

Let u=x2u = x^2u=x2.

Find the corresponding values for yyy.

Split the integral based on the definition of the absolute value.

Rewrite the integral as a sum of two integrals.

Evaluate the first integral.

Evaluate the second integral.

Add the results of the two integrals.

Apply the Weierstrass substitution formulas.

Change the limits of integration.

Substitute these into the integral.

Factor the denominator.

Use partial fraction decomposition.

Evaluate the integral.

Still stuck on this one?

More Mathematics Questions

Differentiate both sides with respect to $x$ using implicit differentiation and the product rule.

Solve for $\frac{dy}{dx}$ by multiplying by $y$ .

Integrate both sides with respect to $x$ to find the equation of the curve.

Use the given point $(1,7)$ to find the constant of integration $C$ .

Let $z = x + iy$ , where $x$ and $y$ are real numbers.

From equation (2), express $y$ in terms of $x$ .

Substitute $y$ into equation (1).

Let $u = x^2$ .

Find the corresponding values for $y$ .