Finding tangent lines and solving logarithmic differential equations

[MarkFL]

Here are the questions:

Find eq of tangent and logarithm differentiation...?

Find an equation of the tangent line to the curve
y=(x^4 -3x^2+2x) * (x^3-2x+3) at x = 0.b:
Use the logarithmic differentiation to differentiate the function

F(x)=(2x+1)^2 * (3x^2-4)^7 *(x+7)^4

I have posted a link there to this topic so the OP can see my work.

 

[MarkFL]

Hello Jawairia,

a) We can save a lot of work by observing we have:

$$y=x^7+\cdots+6x$$

and so, we will find:

$$y'(0)=6$$

We can also easily see that:

$$y(0)=0$$

and so the tangent line must be:

$$y-0=6(x-0)$$

$$y=6x$$

b) We are given:

$$F(x)=(2x+1)^2(3x^2-4)^7(x+7)^4$$

If we take the natural log of both sides, we obtain:

$$\ln(F(x))=\ln\left((2x+1)^2(3x^2-4)^7(x+7)^4 \right)$$

Applying the properties of logarithms, we may write:

$$\ln(F(x))=2\ln(2x+1)+7\ln\left(3x^2-4 \right)+4\ln(x+7)$$

Differentiating with respect to $x$, we find:

$$\frac{1}{F(x)}F'(x)=\frac{4}{2x+1}+\frac{42x}{3x^2-4}+\frac{4}{x+7}$$

Hence:

$$F'(x)=F(x)\left(\frac{4}{2x+1}+\frac{42x}{3x^2-4}+\frac{4}{x+7} \right)$$

Replacing $F(x)$ with its definition, we find:

$$F'(x)=(2x+1)^2(3x^2-4)^7(x+7)^4\left(\frac{4}{2x+1}+\frac{42x}{3x^2-4}+\frac{4}{x+7} \right)$$

Distributing, we get:

$$F'(x)=4(2x+1)(3x^2-4)^7(x+7)^4+42x(2x+1)^2(3x^2-4)^6(x+7)^4+4(2x+1)^2(3x^2-4)^7(x+7)^3$$

Factoring, we find:

$$F'(x)=2(2x+1)(3x^2-4)^6(x+7)^3\left(2(3x^2-4)(x+7)+21x(2x+1)(x+7)+2(2x+1)(3x^2-4) \right)$$

Expanding and collecting like terms, we finally find:

$$F'(x)=2(2x+1)(3x^2-4)^6(x+7)^3\left(60x^3+363x^2+123x-64 \right)$$