Find dy/dx, slope of tangent at x=2 for y = Ln(5x+8)+ex25+9x+3-------.

[sarahlee]

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Section 7 Differentiation 1

7) Given y = L[FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]9[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT]. Find dy/dx. Find the gradient of the tangent that touches the graph at x = 2.
Answer: (Correct to 3 decimal places) The gradient of the tangent is m = .

 

[stapel]

Section 7 Differentiation 1

7) Given y = L[FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]8[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]e[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]9[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]−[/FONT]. Find dy/dx. Find the gradient of the tangent that touches the graph at x = 2.
Answer: (Correct to 3 decimal places) The gradient of the tangent is m = .

I'm sorry, but I don't understand the function expression, as you've posted it...? For instance, what is the meaning of "ex25"?

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!