Primera derivada de funciones - problemas y soluciones

Solución:
$y'=(5x-4)'=(5x)'-(4)'=5-0=5$

Solución:
$y'=(-x^3+4x^2-5)'=(-x^3)'+(4x^2)'+(-5)'=-3x^2+4\times2x+0=-3x^2+8x$

Solución:
$y'=[(x-1)(x^2+3)]'=$

$(x-1)'(x^2+3)+(x-1)(x^2+3)'=$

$(x'-1')(x^2+3)+(x-1)((x^2)'+3')=$

$(1-0)(x^2+3)+(x-1)(2x+0)=$

$(x^2+3)+(x-1)2x=$

$x^2+3+2x^2-2x=3x^2-2x+3$

Solución:
$y'=[(x+2)^2-3x^3]'=[(x+2)^2]'-[3x^3]'=2(x+2)(x+2)'-3\times3\times x^2=$
$2(x+2)(x'+2')-9x^2=2(x+2)(1+0)-9x^2=2(x+2)-9x^2=2x+4-9x^2=-9x^2+2x+4$

Solución:
$y'=\left(\frac{5+x}{x-3}\right)'=\frac{(5+x)'(x-3)-(5+x)(x-3)'}{(x-3)^2}=$

$\frac{(5'+x')(x-3)-(5+x)(x'-3')}{(x-3)^2}=\frac{(0+1)(x-3)-(5+x)(1-0)}{(x-3)^2}=$

$\frac{(x-3)-(5+x)}{(x-3)^2}=\frac{x-3-5-x}{(x-3)^2}=-\frac{8}{(x-3)^2}$

Solución:
$y'=[(x^2-3x+\sqrt{2})^5]'=5(x^2-3x+\sqrt{2})^4(x^2-3x+\sqrt{2})'=$

$5(x^2-3x+\sqrt{2})^4[(x^2)'-(3x)'+(\sqrt{2})']=$

$5(x^2-3x+\sqrt{2})^4(2x-3+0)=5(x^2-3x+\sqrt{2})^4(2x-3)$

Solución:
$y'=\left[\frac{x^2-3x}{x^2+1}\right]'=\frac{(x^2-3x)'(x^2+1)-(x^2-3x)(x^2+1)'}{(x^2+1)^2}=$

$\frac{[(x^2)'-(3x)'](x^2+1)-(x^2-3x)[(x^2)'+(1)']}{(x^2+1)^2}=\frac{(2x-3)(x^2+1)-(x^2-3x)(2x)}{(x^2+1)^2}=$

$\frac{2x^3-3x^2+2x-3-(2x^3-6x^2)}{(x^2+1)^2}=\frac{2x^3-3x^2+2x-3-2x^3+6x^2}{(x^2+1)^2}=$

$\frac{3x^2+2x-3}{(x^2+1)^2}$

Solución:
$y'=(\sqrt{x^2-3})'=\left((x^2-3)^{\frac{1}{2}}\right)'=$

$\frac12(x^2-3)^{\frac{1}{2}-1}(x^2-3)'=\frac12(x^2-3)^{-\frac12}(2x)=$

$\left(\frac{1}{x^2-3}\right)^{\frac12}\cdot x=\frac{x}{(x^2-3)^{\frac12}}=\frac{x}{\sqrt{x^2-3}}=$

$\frac{x}{\sqrt{x^2-3}}\cdot\frac{\sqrt{x^2-3}}{\sqrt{x^2-3}}=\frac{x\sqrt{x^2-3}}{x^2-3}$

Solución:
$y'=\left(\left(\frac{2-x}{x+3}\right)^2\right)'=$

$2\cdot\left(\frac{2-x}{x+3}\right)^{2-1}\cdot\left(\frac{2-x}{x+3}\right)'=$

$2\cdot\left(\frac{2-x}{x+3}\right)^1\cdot\frac{(2-x)'(x+3)-(2-x)(x+3)'}{(x+3)^2}=$

$2\cdot\frac{2-x}{x+3}\cdot\frac{(2'-x')(x+3)-(2-x)(x'+3')}{(x+3)^2}=$

$2\cdot\frac{2-x}{x+3}\cdot\frac{(0-1)(x+3)-(2-x)(1+0)}{(x+3)^2}=$

$2\cdot\frac{2-x}{x+3}\cdot\frac{-(x+3)-(2-x)}{(x+3)^2}=$

$2\cdot\frac{2-x}{x+3}\cdot\frac{-x-3-2+x}{(x+3)^2}=$

$2\cdot\frac{2-x}{x+3}\cdot\frac{-5}{(x+3)^2}=\frac{-10(2-x)}{(x+3)^3}=$

$\frac{10(x-2)}{(x+3)^3}$

Solución:
$y'=(x(x^2-3)(x^2+1))'=x'[(x^2-3)(x^2+1)]+x[(x^2-3)(x^2+1)]'=$

$1\cdot(x^2-3)(x^2+1)+x[(x^2-3)'(x^2+1)+(x^2-3)(x^2+1)']=$

$(x^2-3)(x^2+1)+x[((x^2)'-3')(x^2+1)+(x^2-3)((x^2)'+1')]=$

$(x^2-3)(x^2+1)+x[2x(x^2+1)+(x^2-3)(2x)]=$

$x^4+x^2-3x^2-3+x(2x^3+2x+2x^3-6x)=$

$x^4-2x^2-3+x(4x^3-4x)=$

$x^4-2x^2-3+4x^4-4x^2=5x^4-6x^2-3$

Solución:
$y'=(x^3\cos x)'=(x^3)'\cos x+x^3(\cos x)'=$

$3x^{3-1}\cos x+x^3(-\sin x)=3x^2\cos x-x^3\sin x$

Solución:
$y'=\left(\frac{1-\cos x}{1+\cos x}\right)'=$

$\frac{(1-\cos x)'(1+\cos x)-(1-\cos x)(1+\cos x)'}{(1+\cos x)^2}=$

$\frac{[1'-(\cos x)'](1+\cos x)-(1-\cos x)[1'+(\cos x)']}{(1+\cos x)^2}=$

$\frac{(0+\sin x)(1+\cos x)-(1-\cos x)(0-\sin x)}{(1+\cos x)^2}=$

$\frac{\sin x(1+\cos x)+\sin x(1-\cos x)}{(1+\cos x)^2}=$

$\frac{\sin x(1+\cos x+1-\cos x)}{(1+\cos x)^2}=$

$\frac{2\sin x}{(1+\cos x)^2}$

Solución:
$y'=(\sqrt{7x-x^2-12})'=\frac{1}{2\sqrt{7x-x^2-12}}(7x-x^2-12)'=$

$\frac{1}{2\sqrt{7x-x^2-12}}[(7x)'-(x^2)'-(12)']=$

$\frac{1}{2\sqrt{7x-x^2-12}}(7-2x-0)=$

$\frac{7-2x}{2\sqrt{7x-x^2-12}}$